Log Graphs and Models: 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:

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 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 $T=1296/x^2$.

(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.

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.

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.

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 $k$,

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.

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 $k=\ln(1.785)/4$:

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.

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

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.

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.

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.

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,