Exponentials and Logarithms: past paper questions & worked solutions

At $t=0$, $\log_{10}A=1$, hence $A=10^1=10$ square metres.

Substitute $A=25$ into the model:

\paragraph{Solution} Differentiate $x=e^{2\tan y}$ with respect to $y$:

This is positive on the stated interval, so

Taking logarithms of the original relation gives $\ln x=2\tan y$. Using $\sec^2y=1+\tan^2y$,

Hence $a=2$, $b=4$ and $n=2$.

The gradient in the logarithmic coordinates is

Using the vertical intercept $-2$,

For $x\gt 0$, the logarithm laws give

Therefore

The gradient in the $(x,\log_{10}y)$ coordinates is $(1.5-0)/(0-(-4.8))=1.5/4.8=5/16$. The vertical intercept is $1.5=3/2$, giving the equation below.

Exponentiating base 10 gives

Therefore $k=10^{3/2}=31.62277660\ldots$ and $b=10^{5/16}=2.053525026\ldots$. To three significant figures the model constants are 31.6 and 2.05.

The logarithm is base 10. Thus $10^y=2x+1$ and $2x=10^y-1$.

Differentiate the result of (a) with respect to y:

Take the reciprocal and use $10^y=2x+1$:

The original logarithm requires $x\gt -1/2$.

The horizontal coordinate is $\log_{10}x$ and the vertical coordinate is $\log_{10}y$. The gradient of the line is

The vertical intercept is $4$, so the required equation is

Taking powers of $10$ on both sides gives

Here $x\gt 0$, as required for $\log_{10}x$ to exist. Therefore

At $m=3$,

The predicted heart rate is $191$ beats per minute (3 significant figures).

Using the power law of logarithms,

Therefore

The model, with its coefficient rounded as requested, is

When $m=1$, $h=p$. Thus $p$ is the model's predicted heart rate, in beats per minute, for a mammal of mass $1$ kg.

At $t=0$, $\log_{10}N=2$, so $N=10^2$.

Exponentiating base 10 gives

Thus $a=100$ and $b=10^{0.35}=2.238721\ldots$, which is $2.24$ to 2 decimal places.

Retain the unrounded base $10^{0.35}$ when differentiating:

At $t=5$ this is $0.35\ln(10)\,10^{3.75}=4531.9356\ldots$ bacteria per hour.

Put $X=\log_{10}x$ and $Y=\log_{10}y$. Taking logarithms of $y=10^6/x^3$ gives

At $X=0$, $Y=6$; at $Y=0$, $X=2$. The graph is a straight line of gradient $-3$.

The gradient in the given plot is $(4-0)/(0-(-2))=2$ and its vertical intercept is 4. Therefore $\log_3N=2t+4$. Exponentiating in base 3,

Substitute $t=2$ into the logarithmic model:

Thus $S=30760.968\ldots\ \mathrm{km}^2$. To 3 significant figures, the area is $3.08\times10^4\ \mathrm{km}^2$.

Exponentiating and using index laws,

Thus $p=31622.7766\ldots$ and $q=0.9862795\ldots$. To 3 significant figures, $p=3.16\times10^4$ and $q=0.986$. Hence the complete model, with both constants rounded as requested, is $S=31600(0.986)^t$, where $S$ is in $\mathrm{km}^2$ and $t$ is in years.

The factor $q$ is the yearly multiplier for the sea-floor area covered by coral reefs. Each year, about $98.6\%$ of the previous year's coral-covered sea-floor area remains. This is a yearly decrease of about $1.4\%$ (or $1.372\ldots\%$ using the unrounded value of $q$).

(i) Write $Y=\log_6T$, $X=\log_6x$. The gradient is $(0-4)/(2-0)=-2$ and the intercept is 4, so $Y=4-2X$. (ii) Since $216=6^3$, $\log_6T=4-2(3)=-2$, hence $T=6^{-2}=1/36$.

The log equation gives $\log_6T+2\log_6x=4$. For $x,T\gt 0$, combine logs to obtain $\log_6(Tx^2)=4$. Thus $Tx^2=6^4=1296$, or $T=1296/x^2$.

Since $d(3x^2+k)/dx=6x$, the numerator is $3/2$ times this derivative. The denominator is positive because $k\gt 0$, so

Apply the limits to (a) and combine the logarithms:

Hence $\ln((75+k)/(12+k))=(2/3)\ln8=\ln4$, so $75+k=4(12+k)$. This gives $27=3k$ and $k=9$, consistent with $k\gt 0$.

Raise the power by one, divide by the new power and by the inner derivative 2:

The derivative of $1+2\cos x$ is $-2\sin x$, so a primitive is $-2\ln(1+2\cos x)$. The log argument is positive throughout $0\le x\le\pi/3$. Hence

(i) Extraction starts at $G=0$: $30e^{1-0.05k}=40$, so $1-0.05k=\ln(4/3)$ and $k=20(1-\ln(4/3))=14.24635855\ldots$ years. (ii) This is 14 years and $12(0.24635855\ldots)=2.95630261\ldots$ months after 1 January 1800. Two full months have elapsed from January 1814, so the date is in March 1814. Use the unrounded fractional year to identify the month.

On 1 January 1870, $t=70$. Substitute: $G(70)=40-30e^{1-0.05(70)}=40-30e^{-2.5}=37.53745004\ldots$ tonnes.

As $t\to\infty$, $e^{1-0.05t}\to0$, so $G\to40$ tonnes.

At $r=45$, $\log_{10}M=1.93\log_{10}45+0.684$. Exponentiate to obtain $M=10^{0.684}45^{1.93}=7493.766412\ldots$ kg.

Exponentiate the full equation:

Therefore $p=10^{0.684}=4.83058802\ldots$ and $q=1.93$.

If the base radius is $r=1$ cm, the model gives $M=p\cdot1^q=p$ kg. Thus p is the predicted mass in kilograms of a tree of base radius 1 cm.

Exponentiate in base ten and use index laws:

Therefore $a=10^{1.478}=30.060763\ldots$, which rounds to 30 to the nearest integer, and $b=10^{0.0646}=1.16037937\ldots$, which rounds to 1.16 to three significant figures.

At $t=30$, use the original model to retain precision:

To two significant figures the covered area is $2.6\times10^3\ \mathrm{mm}^2$. Using the rounded model $30(1.16)^{30}$ gives the same two-significant-figure answer.

At the start of the study, $t=0$, so

There are initially $300$ toads.

Set $N=420$ and clear the denominator:

Taking natural logarithms gives

Thus $\boxed{t=16.22\text{ years}}$ to 2 decimal places.

Divide the numerator and denominator by $e^{0.12t}$:

Since $e^{-0.12t}\gt 0$ for every finite $t\geq0$, the denominator is greater than $2$. Hence

The population approaches $450$ as $t$ increases without bound, so it can never reach $500$ according to this model.

Taking base-10 logarithms gives

The vertical intercept is $2.5$ and the gradient is $\dfrac{3.25-2.5}{10-0}=0.075$. Therefore

To 3 significant figures, $\boxed{a=316,\quad b=1.19}$.

Each year the model multiplies the number of recycling households by $b$. This means an increase of approximately $\boxed{19\%\text{ per year}}$ (about $18.85\%$ using the unrounded value).

Retain the unrounded constants:

At $t=5$ this is $129.5021421\ldots$, hence

to 2 significant figures.

Using the chain rule,

Treat $k$ as a constant when integrating:

The stated inequality therefore implies

Since the exponential is increasing and $k+3\gt 0$,

The roots are $(21\pm\sqrt{697})/2$, so the positive upper bound is $k\lt (21+\sqrt{697})/2=23.700378782\ldots$. With $k\gt 1$ an integer, the greatest possible value is $k=23$.

Multiplying the proposed identity by $x-2$ gives

Thus $A=2$, $B-4=-5$ and $C-2(-1)=8$, giving $B=-1$, $C=6$. Hence $g(x)=2x-1+6/(x-2)$, for $x\ne2$.

Since $x-2\gt 0$ throughout $[4,8]$,

Exponentiating in base 10,

Thus $a=10^{1.04}=10.96478196\ldots$ and $b=10^{0.38}=2.398832919\ldots$. To four significant figures the displayed model is $D\approx10.96(2.399)^t$.

Use the original unrounded logarithmic model:

At 12 months,

This exceeds the target of 350000 pounds. The model is increasing because $10^{0.38}\gt 1$, so the target is reached within the first 12 months.

At $t=0$, $18=10+Ae^0=10+A$.

At $t=45$, $16=10+8e^{-45B}$, so $e^{-45B}=6/8=3/4$. Taking logarithms,

Differentiate $T=10+8e^{-Bt}$:

At $t=2$, using the unrounded value $B=\ln(4/3)/45$, $dT/dt=-0.05049372834\ldots$. The negative sign means the temperature is falling.

For every finite $t\ge0$, $8e^{-Bt}\gt 0$ because $B\gt 0$. Thus $T=10+8e^{-Bt}\gt 10$; as $t\to\infty$, T approaches 10 from above. It therefore cannot reach 5 degrees Celsius.

Initially $t=0$ and $N=32$, because $N$ is measured in thousands.

For 100000 phones, use $N=100$:

Taking natural logarithms,

Differentiate the model:

At $t=7$, this is $4.7265337\ldots$ thousand phones per month. Multiplying by 1000 gives $4726.5337\ldots$ phones per month.

For $t\ge0$, $93e^{-0.109t}\gt 0$, so $N\lt 125$ and $N\to125$ as $t\to\infty$. The model can therefore never reach 150000 phones: its limiting count is 125000.

Set $u=x^2+k\gt 0$. Applying the quotient and chain rules,

Comparing with the required form identifies both constants.

The denominator is positive, so stationary points satisfy

There are three distinct real stationary points precisely when $e-k\gt 0$. In that case the factor $1-\ln(x^2+k)$ is positive between the outer roots and negative outside them; $2x$ changes sign at zero. Thus the derivative changes sign at all three points, so all are turning points. Together with $k\gt 0$, the three coordinates below are real and distinct; the restriction is $0\lt k\lt e$.

At $k=e$, the two outer stationary points merge with $x=0$, so there are no longer three distinct turning points. Values of $k$ can approach $e$ from below, but cannot equal it under this condition.

The slope is $(3.85-3.08)/(5-0)=0.154$ and the vertical intercept is 3.08. The plotted vertical coordinate is $\log_{10}N$, not N.

Taking logs of $N=ab^t$ gives $\log_{10}N=\log_{10}a+t\log_{10}b$. Compare with (a): $\log_{10}a=3.08$ and $\log_{10}b=0.154$. Thus $a=10^{3.08}=1202.2644\ldots$ and $b=10^{0.154}=1.4256076\ldots$. To three significant figures:

Use the unrounded logarithmic model to avoid rounding drift:

At $t=0$, the model gives $30=80p/(p+4)$. Thus $30p+120=80p$, so $50p=120$ and $p=2.4\gt 0$.

Put $E=e^{0.15T}$ and use $p=2.4$. At area $50\,\mathrm{m^2}$,

Taking natural logarithms gives $0.15T=\ln(25/9)$, so $T=\ln(25/9)/0.15=6.8110083\ldots$ days.

Divide numerator and denominator by $pe^{0.15t}$: $A=80/(1+(4/p)e^{-0.15t})\longrightarrow80$ as $t\longrightarrow\infty$. Thus the model's limiting pond surface area is $80\,\mathrm{m^2}$; the finite-time covered area approaches this value from below.

Taking base-ten logarithms gives $\log_{10}A=\log_{10}p+t\log_{10}q$. The intercept is 0.32 and the gradient is $(0.56-0.32)/8=0.03$. Thus $p=10^{0.32}=2.08929613\ldots$ and $q=10^{0.03}=1.071519305\ldots$.

Using the parameter values requested in (a), $A=2.089(1.072)^t$. For an arbitrary positive base, $d(q^t)/dt=q^t\ln q$, so

This is a positive area increase per day. Retaining unrounded parameters also gives 0.22 to two decimal places.

Taking logarithms gives $\log_{10}P=\log_{10}a+t\log_{10}b$. Match the intercept and gradient: $\log_{10}a=0.68$, $\log_{10}b=0.09$. Thus $a=10^{0.68}=4.786300923\ldots$ and $b=10^{0.09}=1.230268771\ldots$.

At $t=0$, $P=a$. Thus a is the percentage of the country's population with internet access at the start of 2005, approximately $4.79\%$.

The start of 2015 is $t=10$ years after the start of 2005. Using the model parameters, $P=4.79(1.23)^{10}=37.9652818\ldots\%$, giving $38.0\%$ to three significant figures. Using unrounded parameters gives $10^{0.68+0.09(10)}=38.0189396\ldots\%$, the same rounded answer.

Substitute $f(x)$ into $f$ and multiply numerator and denominator by $x-4$:

The composition requires $x\ne4$ and $f(x)\ne4$, hence also $x\ne16$.

To find $f^{-1}(12)$, solve $f(x)=12$:

Thus $f^{-1}(12)=\frac{32}{5}$.

Put $u=\ln a$. Then $fg(a)=g(a)$ means $f(u)=u$, where $u\ne4$:

Therefore $a=e^8$ or $a=e^{-2}$. Both are positive and neither is $e^4$, so both satisfy the domain restrictions.

At $t=0$, $N=4000/(19+1)=200$ squirrels.

The quotient rule gives

At $t=T$, the derivative is zero. The exponential and denominator are positive, so $19-e^{0.2T}=0$. Therefore $e^{0.2T}=19$ and $A=19$. The derivative is positive before $T$ and negative after $T$, confirming a maximum.

Since $e^{0.2T}=19$, we have $e^{0.1T}=\sqrt{19}$. Substitution into the model gives

The maximum number is $459$ squirrels to the nearest whole number.

At $t=0$, $75=21+A$, so $\boxed{A=54}$.

At $t=5$, $25=21+54e^{-5k}$, hence

Differentiate the temperature model:

A decrease of $9\,{}^{\circ}\mathrm{C}$ per minute means $d\theta/dt=-9$, not $+9$. Thus

Using the unrounded value of $k$,

Since $x\gt 0$, $6\sqrt{x}$ is positive. Therefore $f(x)=0$ requires $\ln(4x)=0$, so $4x=1$ and $x=1/4$.

Apply the product and chain rules:

At $Q$, set $f'(x)=0$. Since $3/\sqrt{x}\gt 0$,

Then $\sqrt{x}=1/(2e)$ and $\ln(4x)=-2$, giving $f(x)=6(1/(2e))(-2)=-6/e$.

The factor $3/\sqrt{x}$ is positive and $\ln(4x)+2$ increases through zero at $Q$, so $f'$ changes from negative to positive: $Q$ is the global minimum. As $x\to\infty$, $f(x)\to\infty$, hence the range of $f$ is $[-6/e,\infty)$. Multiplication by $-2$ reverses the inequality and includes the maximum $12/e$.

At $t=0$, $2500=Ae^0$, so $A=2500$ exactly. At $t=8$, $10000=2500e^{8k}$; hence $e^{8k}=4$ and $k=\ln4/8=0.1732867951\ldots$. To four significant figures:

For the second population,

The requested rate of decrease is the positive magnitude, to three significant figures.

Retain the exact value $k=\ln4/8$. Equating populations gives

Take natural logs and solve:

Exponentiate base 10:

Thus $p=10^{2.74}=549.5408739\ldots$ and $q=10^{0.079}=1.199499303\ldots$. Round p to the nearest integer and q to two significant figures.

At $t=0$, $x=pq^0=p$. Thus p is the initial amount of antibiotic in the patient's bloodstream, approximately 550 milligrams immediately after the dose in this model.

For the second patient, write $x=400e^{-t\ln1.4}$. Differentiate to obtain $dx/dt=-400\ln(1.4)\,1.4^{-t}$. At $t=5$, the derivative is $-25.02470969\ldots\,\mathrm{mg\,h^{-1}}$. The negative sign indicates the amount is decreasing.

Write the integrand as $12(2x-1)^{-2}$. Reverse the chain rule, dividing by the inner derivative 2:

(a) Since $4x+3=4(x+2)-5$,

(b) A primitive is $4x-5\ln|x+2|$. The absolute value is essential here: $x+2$ is negative throughout $[-8,-5]$.

First evaluate the inner function:

Then apply $f$ to this result:

Put $y=f(x)$ and rearrange to make $x$ the subject:

For $x\gt 0$, the denominator $x+1\gt 1$, so $0\lt f(x)\lt 12$. Conversely, every $y$ with $0\lt y\lt 12$ gives $x=12/y-1\gt 0$. Thus the range of $f$, and hence the domain of its inverse, is $(0,12)$. Interchanging $x$ and $y$ gives

Both functions must be defined, so $0\lt x\lt 12$. Substitute the formula for the inverse and multiply by the nonzero quantity $x(x+1)$:

The algebraic candidates are $x=-4$ and $x=3$. Reject $x=-4$ because it lies outside $0\lt x\lt 12$. For $x=3$, $f^{-1}(3)=12/3-1=3=f(3)$, so the only real solution is

Since $e^{2x}\gt 0$ for every real $x$, the range is $\boxed{(3,\infty)}$.

Put $y=\sqrt{x+5}$. The original domain is $x\gt -5$, so $y\gt 0$. Squaring gives $y^2=x+5$, hence $x=y^2-5$. Interchanging the input and output,

Evaluate the inside function first:

Because $a\gt 0$, $g(a^2-5)=\sqrt{a^2}=a$, while $g(59)=8$. Thus

Here $p=\sqrt5/2$ is irrational and greater than 1, so the answer satisfies $a\gt 0$.

At $t=5$, the growth rate is $180e=489.290729\ldots$ insects per year.

At $t=0$, $10000=8000+P$, so $P=2000$ exactly. Using the population at $t=4$,

Taking natural logarithms,

At equality, $8000+900e^{0.2T}=8000+2000e^{kT}$, hence

Use the unrounded value $k=\ln(1.785)/4$:

Since $x^2+3\ge3$, with equality at $x=0$, and the logarithm is increasing, the range of $f$ is $[\ln3,\infty)$.

Let $y=(3+5x)/(x+2)$. Then

Moreover, $g(x)=5-7/(x+2)$ with $x\gt -2$ has range $(-\infty,5)$. Thus

Therefore

Taking natural logarithms gives $a=\frac12\ln5$. The input $e^{2a}=5$ satisfies the domain of $g$.

The gradient is $(2.25-2)/(5-0)=0.05$ and the vertical intercept is $2$. Hence $\log_{10}V=0.05t+2$.

Thus $a=100$ exactly and $b=10^{0.05}=1.122018454\ldots=1.12$ to three significant figures. In the requested rounded-parameter form, $V=100(1.12)^t$; retain the unrounded base for later calculations.

Differentiating the unrounded model,

At $t=T$ this equals $50$, so

Therefore $T=13$ years to the nearest integer.

Multiply the identity by $x^2+4$:

Comparing the cubic, quadratic and linear coefficients gives $A=4$, $B=2$ and $4A+C=3$, hence $C=3-16=-13$.

The $x^2$ coefficients give $B=2$, while the constant terms give $4B+D=8$. Substitute $B=2$: $4(2)+D=8$, so $D=0$.

Using the decomposition, the integrand is $4x+2-13x/(x^2+4)$. Since the derivative of $x^2+4$ is $2x$, an antiderivative is

Therefore

Combine the logarithms: $36-\frac{13}{2}\ln4=36-13\ln2$.

At $t=0$, $e^{-0.05t}=1$, so $H=280+24=304$ degrees Celsius.

The curve starts at $(0,304)$ and decreases for $t\ge0$. The exponential term stays positive and tends to zero, so the curve approaches $H=24$ from above without crossing it.

Substitute $H=144$:

Take natural logarithms: $-0.05t=\ln(3/7)$. Hence $t=-20\ln(3/7)=20\ln(7/3)=16.9459572077\ldots$ minutes.

Differentiate using the chain rule:

The original equation gives $280e^{-0.05t}=H-24$, so

The logarithm requires $x+1\gt 0$, so the domain is $x\gt -1$. As $x\to-1^+$, $\ln(x+1)\to-\infty$ and the ordinate tends to infinity.

(i) At the $y$-axis, $x=0$, giving $y=|2-4\ln1|=2$.

(ii) At the $x$-axis, $|2-4\ln(x+1)|=0$, so

Thus $A=(0,2)$ and $B=(e^{1/2}-1,0)$.

On $x\gt -1$, split the absolute-value inequality into two alternatives:

Dividing by $-4$ reverses the inequality in each branch:

Since the exponential is increasing, this becomes $x+1\lt e^{-1/4}$ or $x+1\gt e^{5/4}$. Intersect with the logarithmic domain; neither boundary is included because the source inequality is strict.

At the start $t=0$, so $e^{kt}=1$ and the model gives $F(0)=350/(9+1)=35$ fruit flies.

Using $F(15)=200$,

Take natural logs to obtain $15k=\ln12$, as required.

By the quotient rule,

Set $F'(T)=10$ and let $u=e^{kT}\gt 0$. Then

With the exact $k=\ln12/15$, the quadratic formula gives

Both values are positive and at least 1, hence both correspond to $T\ge0$. Finally $T=\ln u/k$ gives $5.6779528361\ldots$ or $20.8488936858\ldots$ days. To one decimal place the two times are as follows.

Clear the denominator in the proposed identity:

Matching coefficients gives $A=2$, $3A+B=0$, hence $B=-6$. Then $4A+3B+2C=-4$ gives $C=3$; the constant checks as $4B+3C=-24+9=-15$. Therefore $f(x)=2x-6+3(2x+3)/(x^2+3x+4)$.

The numerator $2x+3$ is the derivative of $x^2+3x+4$. This quadratic is $(x+3/2)^2+7/4\gt 0$, so

Using both limits,

Thus $p=4$ and $q=8$ are integers, as required.

At P, $18=5e^{x-1}+3$, so $e^{x-1}=3$. Taking natural logarithms gives $x-1=\ln3$, hence $x=1+\ln3=\ln e+\ln3=\ln(3e)$.

Define $F(x)=5e^{x-1}+x^2-7$; its zeros are the intersection abscissae, and F is continuous. At the endpoints of the three-decimal rounding interval,

Continuity guarantees a root between them. Moreover $F'(x)=5e^{x-1}+2x\gt 0$ on this positive interval, so that root is unique there. It is the positive intersection $\alpha$, and rounds to 1.134 to three decimal places.

Start at $x_1=-3$ and keep full precision in $x_{n+1}=-\sqrt{7-5e^{x_n-1}}$:

The later iterates agree to six decimal places. The negative root corresponds to $\beta$ in the figure.

By the product and chain rules,

For $-\pi/4\lt x\lt \pi/4$, $e^{3x}\sec2x$ is positive and nonzero. Hence $g'(x)=0$ precisely when $3+2\tan2x=0$, or $\tan2x=-3/2$. Since $-\pi/2\lt 2x\lt \pi/2$, tangent is one-to-one there:

Differentiate $x=\ln(\sin y)$ with respect to y:

The original equation gives $\sin y=e^x$. Since $0\lt y\lt \pi/2$, cosine is positive, so $\cos y=\sqrt{1-\sin^2y}=\sqrt{1-e^{2x}}$. Therefore

Here $x\lt 0$ from the original interval, so the square root is real and nonzero.

Increase the power to $8$ and divide by both $8$ and the inner derivative $3$:

The denominator is $x^2+x-2$. The first quotient term is $4x$. Subtracting $4x(x^2+x-2)$ leaves $-3x^2+x+14$. The next quotient term is $-3$. Subtracting $-3(x^2+x-2)$ leaves $4x+8$. Thus polynomial division gives

Since $4x+8=4(x+2)$,

Hence $a=4$, $b=-3$, $c=4$.

For $x\gt 1$, an antiderivative is $2x^2-3x+4\ln(x-1)$, so

Thus the exact integral is $126+8\ln2$.

To find $f^{-1}(22)$, solve $f(u)=22$: $2+5\ln u=22$, so $\ln u=4$ and $u=e^4\gt 0$.

Using the quotient rule,

For every $x\gt 1/3$, the denominator is positive, so $g'(x)\gt 0$. Thus g is strictly increasing on its entire domain.

Set $y=(6x-2)/(2x+1)$. Then

As $x\downarrow1/3$, $g(x)\to0$; as $x\to\infty$, $g(x)\to3$. Since g is strictly increasing, its range is $(0,3)$; neither limiting value is attained. Interchange the input and output symbols.

The inner function g takes every value in $(0,3)$. Since $f(u)=2+5\ln u$ is increasing for $u\gt 0$, its values on this interval run from $-\infty$ to, but not including, $2+5\ln3$.

At the first landing, $h=0$ and $x=d\gt 0$:

Taking logarithms, $0.02d=\ln3$, hence $d=50\ln3=54.930614\ldots$. The root $x=0$ is the launch point, not the landing point.

By the product and chain rules,

At a stationary point,

Since $x+50\gt 0$, logarithms give $0.02x=\ln(150/(x+50))$. Also $h”(x)=-(0.02+0.0002x)e^{0.02x}\lt 0$ on $[0,d]$. Thus the stationary point is the unique maximum; the endpoints both have zero height.

(i) With $x_1=30$,

(ii) Continuing with full intermediate precision gives $x_3=30.54431108\ldots$, $x_4=31.09139065\ldots$, $x_5=30.75292479\ldots$ and $x_6=30.96205562\ldots$. Continued iteration converges to $30.88212\ldots$, which lies in $(0,d)$. This is the horizontal distance at maximum height, not the maximum height itself.

Since $3x-1\gt 0$ throughout $3\le x\le42$, an antiderivative is

The factor $1/3$ compensates for the derivative of $3x-1$. Thus

Using $\ln(a^3)=3\ln a$ for $a\gt 0$, the exact value in simplest form is

Multiply the given decomposition by $(x-1)^2$:

Comparing coefficients gives

The coefficient of $x$ also agrees, since $A-2B=2-2(-3)=8$. Hence

Integrating each term,

Therefore, for $x\gt 1$,

where $K$ is an arbitrary constant of integration, distinct from the coefficient $C=4$.

At one year,

The negative sign indicates a loss. Its magnitude is about $828058$ pounds, which is approximately $830000$ pounds.

The model is continuous for $t\ge0$. Evaluate

Continuity and the sign change give a zero between 6 and 7.

Set $P=0$ and move the logarithm to the other side:

Multiply by 10 and divide by 4 to obtain the required form.

Starting with $t_1=6$, retain full calculator precision at each substitution:

Therefore the requested values to three decimal places are as follows.

The break-even time is approximately $6.315$ years, or $6.315\times12=75.78$ months. Profit starts just after that time; the first complete number of months is therefore 76, not 75. Indeed $P(75/12)\lt 0$ and $P(76/12)\gt 0$.

At $t=0$, $18=A-180e^0=A-180$. Hence $A=198$.

At $t=5$, $90=198-180e^{-5k}$, so $180e^{-5k}=108$ and $e^{-5k}=3/5$. Taking logarithms gives $-5k=\ln(3/5)$, hence $k=(1/5)\ln(5/3)$. Thus one suitable rational pair is $p=1/5$, $q=5/3$.

Use the unrounded k:

Round the temperature to three significant figures.

Differentiate with respect to time: $d\theta/dt=180ke^{-kt}$. At $t=9$ this is $180[\ln(5/3)/5]e^{-(9/5)\ln(5/3)}=7.332421555\ldots$. The sign is positive because the oven is warming.

Prices in the model are in thousands of pounds, so

Taking logarithms gives $T=10\ln2=6.9314718\ldots$, hence $\boxed{T=6.93\text{ years}}$.

Thus the answer is $\boxed{-0.97\text{ thousand pounds per year}}$ to 2 decimal places.

At $t=0$, $36=4+m$, so $\boxed{m=32}$.

The initial car price is $7+16=23$ thousand pounds. The difference is $36-23=13$ thousand pounds, or $\boxed{\pounds13\,000}$.

Equating the models with $m=32$ gives

Let $u=e^{-0.1\alpha}\gt 0$. Then $32u^2-16u-3=0$, so

Reject the negative root because an exponential is positive. Hence

This value is inside the model interval $0\leq t\leq15$.

At $t=0$, $H=32+40-20=52$.

As $t\to\infty$, both decaying exponential terms tend to zero.

At the maximum,

Also $H”(t)=1.6e^{-0.2t}-16.2e^{-0.9t}$. At the stationary point, $e^{-0.9T}=(4/9)e^{-0.2T}$, so $H”(T)=-5.6e^{-0.2T}\lt 0$, confirming a maximum.

For $H=37$,

Taking logarithms gives

Using $-\ln u=\ln(1/u)$ for $u\gt 0$ then gives

Starting with $t_1=10$,

Continuing without rounding intermediate values gives $t_3=10.3954778346\ldots$, $t_4=10.3954789826\ldots$ and $t_5=10.3954789844\ldots$. Thus (i) $t_2=10.3947$, and (ii) $M=10.3955$ minutes, both to 4 decimal places.

Write $y=(16/9)(3x-k)^{-1}$. The chain rule gives

At $x=1$, the given gradient yields

Hence $3-k=\pm2/3$. Both resulting k-values are positive and different from 3.

Since $k\lt 3$, use $k=7/3$. At $x=1$, $y=16/[9(3-7/3)]=8/3$. The tangent gradient is $-12$, so the normal gradient is $1/12$:

With $k=7/3$, the denominator does not vanish on $[1,3]$. Using the reciprocal-linear antiderivative,

Differentiate the velocity:

At a stationary velocity, $e^{t-10}=9e^{-0.75t}$, so $e^{1.75t}=9e^{10}$ and $t=(10+\ln9)/1.75=6.9698426\ldots$ seconds. The second derivative $-e^{t-10}-6.75e^{-0.75t}$ is negative, so this is a maximum. Substitute this unrounded time into v to get $11.88728125\ldots$ metres per second.

Integrating the velocity gives $12t-e^{t-10}+16e^{-0.75t}$. Evaluate between 0 and T:

Rearrange to $12T=116-16e^{-0.75T}+e^{T-10}-e^{-10}$ and divide by 12.

With $T_1=10$, substitution gives $T_2=9.7492587708\ldots$. Continuing with unrounded values gives

Further iteration settles at $9.7293319307\ldots$ seconds. Thus (i) $T_2=9.7493$ and (ii) the race time is $9.7293$ seconds, to four decimal places.

Factor the first denominator:

On the stated domain $x\gt 2$, both denominators and $x+4$ are nonzero. Hence

The quotient rule gives

For every $x\gt 2$, the squared denominator is positive; hence $f'(x)\lt 0$. Therefore f is strictly decreasing on its whole domain.

Set $y=3+2\ln x$. Then $\ln x=(y-3)/2$, so $x=e^{(y-3)/2}$. For $x\ge1$, the original range is $y\ge3$. Interchanging the input and output symbols gives the inverse with its domain.

The equation $g(f(a))=5$ gives $3+2\ln(f(a))=5$, so $\ln(f(a))=1$ and $f(a)=e$. Using the simplified f,

The resulting value is approximately $2.2083\gt 2$, and the inner output $e\ge1$ is admissible for g.

At the start, $t=0$ and $e^{0.3t}=1$. Thus $N(0)=600/(2+1)=200$.

Divide numerator and denominator by $e^{0.3t}$ to write $N=600/(1+2e^{-0.3t})$. As $t\to\infty$, the denominator tends to 1, so the upper limit is 600 fish.

Set $N=500$ and multiply through: $1000+500e^{0.3t}=600e^{0.3t}$, so $e^{0.3t}=10$. Hence $t=\ln10/0.3=7.675283643\ldots$ years. The fractional year is $12(0.675283643\ldots)=8.10340372\ldots$ months. Round only the months to the nearest whole month.

The quotient rule gives

The two $e^{0.6t}$ terms cancel exactly.

Let $z=e^{0.3T}$. Since $T\ge0$, require $z\ge1$. Then $360z/(2+z)^2=8$ gives $45z=(2+z)^2$, hence

The smaller root lies between 0 and 1 and would give $T\lt 0$, so reject it. For the larger root,

Factor the denominator as $(x-4)(x+3)$ and multiply the proposed identity through by it:

The right side expands to $x^4-x^3+(P-12)x^2+(Q-P)x+3Q-12P$. Comparing coefficients gives $P-12=-10$, so $P=2$, and $Q-P=3$, so $Q=5$. The constant check is $3(5)-12(2)=-9$, as required. The identity holds where the original fraction is defined: the stated $x\gt -3$ still excludes $x=4$.

From (a), $g(x)=x^2+2+5/(x-4)$ on the curve domain. Thus $g'(x)=2x-5/(x-4)^2$. At $x=2$, $g(2)=4+2-5/2=7/2$ and $g'(2)=4-5/4=11/4$. The tangent through $(2,7/2)$ is

The source shading is above the x-axis from $x=0$ to $x=2$, so integrate g over those bounds.

The absolute value is essential because $x-4\lt 0$ throughout this interval. Evaluating gives

Thus $a=20/3$ and $b=-5$ in the requested form.