Exponentials and Logarithms: past paper questions & worked solutions

The logarithms require $12-2x>0$ and $x+1>0$, so $-1<x<6$.

$2=\log_4(16)$, so $2+2\log_4(x+1)=\log_4\left(16(x+1)^2\right)$.

$\log_4(12-2x)=\log_4\left(16(x+1)^2\right)$ gives $12-2x=16(x+1)^2$.

$12-2x=16x^2+32x+16$, so $8x^2+17x+2=0$.

$8x^2+17x+2=(8x+1)(x+2)=0$, so $x=-\dfrac18$ or $x=-2$.

The value $x=-2$ is outside $-1<x<6$, so the solution is $x=-\dfrac18$.

The logarithms require $16x>0$, $x+1>0$ and $x+6>0$, so $x>0$.

$\log_2(16x)+\log_2(x+1)=\log_2(16x(x+1))$.

$3=\log_2 8$, so $3+\log_2(x+6)=\log_2(8(x+6))$.

Therefore $16x(x+1)=8(x+6)$.

$16x^2+16x=8x+48$, so $16x^2+8x-48=0$. Dividing by $8$ gives $2x^2+x-6=0$.

$(2x-3)(x+2)=0$, so $x=\dfrac32$ or $x=-2$. The domain $x>0$ rejects $-2$, leaving $x=\dfrac32$.

$2\log_3(1-x)=\log_3(1-x)^2$ and $3=\log_3 27$.

The equation becomes $\log_3(32-12x)=\log_3\{27(1-x)^2\}$.

$32-12x=27(1-x)^2$.

$32-12x=27-54x+27x^2$, so $27x^2-42x-5=0$.

$x=\dfrac{42\pm48}{54}$, giving $x=\dfrac53$ or $x=-\dfrac19$.

The original logarithms require $1-x>0$, so $x<1$. Therefore $x=\dfrac53$ is invalid and $x=-\dfrac19$.

The logarithms require $2-x>0$ and $x+10>0$, so $-10<x<2$.

$2\log_2(2-x)=\log_2((2-x)^2)$.

$4=\log_2 16$, so $4+\log_2(x+10)=\log_2(16(x+10))$.

$(2-x)^2=16(x+10)$.

$x^2-4x+4=16x+160$, so $x^2-20x-156=0$.

$x=26$ or $x=-6$. The value $26$ is outside the domain, so $x=-6$.

$\log_{\sqrt a}(a^6)=\log_{a^{1/2}}(a^6)=\dfrac{6}{1/2}=12$.

$3^a=70$ gives $a=\log_3 70=\dfrac{\log 70}{\log 3}$.

$a=3.867147\ldots$, so $a=3.867$ to 3 decimal places.

$4=\log_3(3^4)=\log_3 81$.

$4+3\log_3 b=\log_3 81+\log_3(b^3)=\log_3(81b^3)$.

The equation becomes $\log_3(81b^3)=\log_3(5b)$, so $81b^3=5b$.

Since $\log_3 b$ is defined, $b>0$. Divide by $b$ to get $81b^2=5$.

$b^2=\dfrac{5}{81}$, and $b>0$, so $b=\dfrac{\sqrt5}{9}$.

Since $2=\log_3 9$, the equation becomes $\log_3(4x)+\log_3 9=\log_3(5x+7)$.

$\log_3(4x)+\log_3 9=\log_3(36x)$.

Therefore $36x=5x+7$, so $31x=7$ and $x=\dfrac{7}{31}$.

Let $L=\log_a y$. Then $\log_a(y^r)=r\log_a y=rL$.

$\displaystyle \sum_{r=1}^{2}\log_a(y^r)=L+2L=3L$.

$\displaystyle \sum_{r=1}^{2}(\log_a y)^r=L+L^2$.

So $3L=L+L^2$, which gives $L^2-2L=0$ and $L(L-2)=0$.

Since $a>1$ and $y>1$, $\log_a y>0$, so $L\ne0$. Hence $L=2$ and $y=a^2$.

$\log_4 a+\log_4 b=\log_4(ab)=3$.

Therefore $ab=4^3=64$.

From $a-b=8$, $b=a-8$.

$a(a-8)=64$, so $a^2-8a-64=0$.

$a=\dfrac{8\pm\sqrt{64+256}}{2}=4\pm4\sqrt5$.

Since $b=a-8$ must also be positive, take $a=4+4\sqrt5$.

$b=a-8=4\sqrt5-4$.

Since $P(k,300000)$ lies on $y=3\times2^{-x}$, $300000=3\times2^{-k}$.

$100000=2^{-k}$, so $-k=\log_2(100000)$ and $k=-\log_2(100000)=-16.609\ldots$.

To $2$ decimal places, $k=-16.61$.

The $x$ values increase by $1.5$, so $h=1.5$.

$\displaystyle\int_{-0.5}^{7}3\times2^{-x}dx\approx\dfrac{1.5}{2}\{4.243+0.023+2(1.5+0.530+0.188+0.066)\}$.

This gives $6.6255$, which is $6.63$ to $2$ decimal places.

$\displaystyle\int_{-0.5}^{7}2^{-x}dx=\dfrac13\int_{-0.5}^{7}3\times2^{-x}dx$.

Using the substitution $u=-x$, $\displaystyle\int_{-7}^{0.5}2^x dx=\int_{-0.5}^{7}2^{-u}du$.

The required estimate is therefore $\dfrac23(6.6255)=4.417$, which is $4.42$ to $2$ decimal places.

$\log_2 9=\log_2(3^2)=2\log_2 3=2a$.

$\log_2\left(\dfrac{\sqrt3}{16}\right)=\log_2(3^{1/2})-\log_2(16)$.

$\log_2(3^{1/2})=\dfrac12\log_2 3=\dfrac a2$ and $\log_2(16)=4$, so the expression is $\dfrac a2-4$.

For $3^x\times2^{x+4}=6$, take $\log_2$ of both sides: $\log_2(3^x)+\log_2(2^{x+4})=\log_2 6$.

$\log_2(3^x)=xa$, $\log_2(2^{x+4})=x+4$, and $\log_2 6=\log_2(2\cdot3)=1+a$.

So $xa+x+4=a+1$. Hence $x(a+1)=a-3$, giving $x=\dfrac{a-3}{a+1}$.

The strip width is $h=3$. Therefore $\int_2^{14}\log_2(2x)\,dx\approx\dfrac32\{2+4.81+2(3.32+4+4.46)\}$.

The estimate is $\dfrac32(30.37)=45.555$, so to one decimal place the answer is $45.6$.

$\log_2(4x^2)=\log_2\{(2x)^2\}=2\log_2(2x)$, so $\int_2^{14}\dfrac{\log_2(4x^2)}5\,dx=\dfrac25\int_2^{14}\log_2(2x)\,dx$.

Using part (a), the estimate is $\dfrac25(45.555)=18.222$, so $18.2$ to one decimal place.

$\log_2\left(\dfrac2x\right)=2-\log_2(2x)$.

$\int_2^{14}\log_2\left(\dfrac2x\right)dx\approx\int_2^{14}2\,dx-45.555=24-45.555=-21.555$, so $-21.6$.

$y=3\times4^x$ is an increasing exponential curve. At $x=0$, $y=3$, so the curve crosses the $y$-axis at $(0,3)$ and has no $x$-axis intersection.

At $P$, $6^{1-x}=3\times4^x$.

$(1-x)\log_{10}6=\log_{10}3+x\log_{10}4$.

$\log_{10}6-\log_{10}3=x(\log_{10}4+\log_{10}6)$.

$\log_{10}6-\log_{10}3=\log_{10}2$ and $\log_{10}4+\log_{10}6=\log_{10}24$.

Therefore $x=\dfrac{\log_{10}2}{\log_{10}24}$.

Since $\log_a b=k$, we can replace every $\log_a b$ by $k$.

$\log_a\left(\dfrac{\sqrt a}{b}\right)=\log_a(a^{1/2})-\log_a b=\dfrac12-k$.

$\log_a(a^2b)=\log_a a^2+\log_a b=2+k$, and $\log_a(b^3)=3\log_a b=3k$.

Therefore $\dfrac{\log_a(a^2b)}{\log_a(b^3)}=\dfrac{k+2}{3k}$.

$\log_a(b^n)=n\log_a b=nk$, so $k+\log_a(b^n)=k+nk=(n+1)k$.

$\sum_{n=1}^{50}(n+1)k=k\left(\sum_{n=1}^{50}n+50\right)=k(1275+50)=1325k$.

The width is $1$, so $\displaystyle\int_{-1}^{4}\log_{10}(x+5)\,dx\approx\dfrac12\{\log_{10}4+\log_{10}9+2(\log_{10}5+\log_{10}6+\log_{10}7+\log_{10}8)\}$.

This is $\log_{10}\left(\sqrt{4\times9}\times5\times6\times7\times8\right)$.

$\sqrt{4\times9}\times5\times6\times7\times8=6\times1680=10080$, so $k=10080$.

The logarithms require $5-a>0$ and $a+25>0$, so $-25<a<5$.

$2\log_5(5-a)-\log_5(a+25)=\log_5\left(\dfrac{(5-a)^2}{a+25}\right)$.

$1=\log_5 5$, so $\dfrac{(5-a)^2}{a+25}=5$.

$(5-a)^2=5(a+25)$, so $a^2-10a+25=5a+125$. Hence $a^2-15a-100=0$.

$(a-20)(a+5)=0$, so $a=20$ or $a=-5$. Only $a=-5$ is in the domain, so $a=-5$.

The dormice model is $D=2000(1.03)^t$.

$D=2000(1.03)^6=2388.10\ldots$, so there are $2390$ dormice to 3 significant figures.

Let the vole model be $N=ab^t$.

At $t=4$, $ab^4=3690$. At $t=7$, $ab^7=3470$.

Dividing gives $b^3=\dfrac{3470}{3690}$, so $b=0.9797\ldots$, which is $0.98$ to 2 significant figures.

$a=\dfrac{3690}{b^4}=4005.18\ldots$, which is $4000$ to 2 significant figures.

So the model is approximately $N=4000(0.98)^t$.

Use $2000(1.03)^T=a b^T$ with the unrounded constants from the vole model.

$\left(\dfrac{1.03}{b}\right)^T=\dfrac{a}{2000}$, so $T=\dfrac{\log(a/2000)}{\log(1.03/b)}$.

Substituting $a=4005.18\ldots$ and $b=0.9797\ldots$ gives $T=13.875\ldots$, so $T=13.9$ to one decimal place.

$2\log_4(x+3)=\log_4((x+3)^2)$, so the left side is $\log_4(x(x+3)^2)$.

$\dfrac12=\log_4 2$, so the right side is $\log_4(4x+2)+\log_4 2=\log_4(8x+4)$.

Therefore $x(x+3)^2=8x+4$.

$x(x^2+6x+9)=8x+4$, so $x^3+6x^2+9x=8x+4$.

Rearranging gives $x^3+6x^2+x-4=0$.

Since $-1$ is a root, $(x+1)$ is a factor of $x^3+6x^2+x-4$.

$x^3+6x^2+x-4=(x+1)(x^2+5x-4)$.

$x^2+5x-4=0$ gives $x=\dfrac{-5\pm\sqrt{25+16}}{2}=\dfrac{-5\pm\sqrt{41}}{2}$.

The original log equation requires $x>0$. The only positive root is $x=\dfrac{-5+\sqrt{41}}{2}$.

When $t=3$, $w=10$. So $10=\log_a(3+5)-\log_a4=\log_a8-\log_a4$.

$\log_a8-\log_a4=\log_a\left(\dfrac84\right)=\log_a2$, so $10=\log_a2$.

$10=\log_a2$ means $a^{10}=2$, hence $a=2^{1/10}=1.071773\ldots=1.072$ to 3 decimal places.

Using $a=1.072$, $w=\log_{1.072}(t+5)-\log_{1.072}4$.

$w=\log_{1.072}\left(\dfrac{t+5}{4}\right)$.

Changing from logarithmic form to exponential form gives $(1.072)^w=\dfrac{t+5}{4}$, so $t=4(1.072)^w-5$.

When $w=15$, $t=4(1.072)^{15}-5=6.3496\ldots$, so $t=6.35$ months to 3 significant figures.

Using $y=1-\log_{10}(\sin x)$ gives $y(1)=1.075$ and $y(2)=1.041$ to 3 decimal places.

The $x$ values increase by $0.5$, so $h=0.5$.

$A\approx\dfrac{0.5}{2}\{1.319+1.850+2(1.075+1.001+1.041+1.223)\}$.

This gives $2.9625\ldots$, so the estimate is $2.96$ to 2 decimal places.

Since $y=1-\log_{10}(\sin x)$, we have $\log_{10}(\sin x)=1-y$.

$3+\log_{10}(\sin x)=4-y$, and the interval from $0.5$ to $3$ has length $2.5$.

$\int_{0.5}^{3}(3+\log_{10}(\sin x))\,dx\approx4(2.5)-2.9625\ldots=7.0374\ldots$, so $7.04$.

Four equal strips from $x=2$ to $x=14$ have width $h=\dfrac{14-2}{4}=3$.

Use $x=2,5,8,11,14$ and $y=\log_{10}x$.

$A\approx\dfrac32\{\log_{10}2+\log_{10}14+2(\log_{10}5+\log_{10}8+\log_{10}11)\}=10.10\ldots$.

Use more strips of smaller width to make the trapezium rule estimate more accurate.

$\log_{10}\sqrt{x}=\dfrac12\log_{10}x$, so $\int_2^{14}\log_{10}\sqrt{x}\,dx\approx\dfrac12(10.10)=5.05$.

$\log_{10}(100x^3)=\log_{10}100+\log_{10}x^3=2+3\log_{10}x$.

$\int_2^{14}(2+3\log_{10}x)\,dx\approx2(12)+3(10.10)=54.30$.

$y=2^x$ is an increasing exponential curve. It crosses the $y$-axis at $(0,1)$ and has no $x$-axis intercept.

$y=4^x-6$ is the graph of $y=4^x$ translated $6$ units down. It crosses the $y$-axis at $(0,-5)$.

For $y=4^x-6$, set $4^x-6=0$. Hence $4^x=6$ and $x=\log_4 6$.

At $P$, $2^x=4^x-6$.

Let $u=2^x$. Then $4^x=(2^x)^2=u^2$, so $u=u^2-6$.

$u^2-u-6=0$, so $(u-3)(u+2)=0$. Since $u=2^x>0$, $u=3$.

$2^x=3$, so $x=\log_2 3$, and $y=3$. Therefore $P=(\log_2 3,3)$.

From $7^{x+2}=3$, divide by $7^2$ to get $7^x=\dfrac3{49}$.

Therefore $x=\log_7\left(\dfrac3{49}\right)$.

Since $1=\log_2 2$, the equation becomes $\log_2 2+\log_2 y+\log_2(y+4)=\log_2(5-y)$.

$\log_2\{2y(y+4)\}=\log_2(5-y)$.

Thus $2y(y+4)=5-y$, so $2y^2+9y-5=0=(2y-1)(y+5)$.

The original logarithms require $y>0$ and $y<5$, so $y=-5$ is invalid and $y=\dfrac12$.

Let the yearly ratio be $r$. After 3 years, $30000r^3=34000$.

$r^3=\dfrac{17}{15}$, so $r=\left(\dfrac{17}{15}\right)^{1/3}\approx1.04257$. Therefore $p=(r-1)\times100\approx4.26$.

After $N$ years the model gives $30000r^N$. We need $30000r^N>75000$, so $r^N>2.5$.

$N\log r>\log(2.5)$, so $N>\dfrac{\log(2.5)}{\log r}$.

Using $r\approx1.04257$, $\dfrac{\log(2.5)}{\log r}\approx21.98$. Hence the smallest integer value is $N=22$.

From $8^{2x-5}=20$, take $\log_2$ of both sides to get $(2x-5)\log_2 8=\log_2 20$.

Since $\log_2 8=3$ and $\log_2 20=\log_2(4\cdot5)=2+\log_2 5$, we get $3(2x-5)=2+\log_2 5$.

$6x-15=2+\log_2 5$, so $6x=17+\log_2 5$ and $x=\dfrac{17}{6}+\dfrac16\log_2 5$.

$3=\log_3 27$ and $2\log_3(4-y)=\log_3(4-y)^2$.

$\log_3(13+2y)+\log_3 27=\log_3(4-y)^2$, so $27(13+2y)=(4-y)^2$.

$351+54y=y^2-8y+16$, hence $y^2-62y-335=0=(y-67)(y+5)$.

We need $13+2y>0$ and $4-y>0$. Therefore $y=67$ is not valid, and the solution is $y=-5$.

Substitute $t=4$: $3\log_2(8)-2\log_2(2)=3(3)-2(1)=7$.

$3\log_2(t+4)-2\log_2(t-2)=\log_2\left(\dfrac{(t+4)^3}{(t-2)^2}\right)$.

$7=\log_2(128)$, so $\dfrac{(t+4)^3}{(t-2)^2}=128$.

$(t+4)^3=128(t-2)^2$.

$t^3+12t^2+48t+64=128t^2-512t+512$.

Rearranging gives $t^3-116t^2+560t-448=0$.

Since $t=4$ is a root, $(t-4)$ is a factor.

$t^3-116t^2+560t-448=(t-4)(t^2-112t+112)$.

$t^2-112t+112=0$ gives $t=\dfrac{112\pm\sqrt{112^2-4(112)}}{2}=56\pm12\sqrt{21}$.

The original equation requires $t>2$. The value $56-12\sqrt{21}$ is less than $2$, so it is rejected. Hence $t=4$ or $t=56+12\sqrt{21}$.

For $k=1$, $N=\lambda^t$ with $0<\lambda<1$. The graph starts at $(0,1)$, decreases as $t$ increases, and tends towards $0$.

After $5700$ years, $N=\dfrac{k}{2}$, so $k\lambda^{5700}=\dfrac{k}{2}$.

$\lambda^{5700}=\dfrac12$, hence $\lambda=\left(\dfrac12\right)^{1/5700}=0.999878\ldots$, so $\lambda=0.999878$ to 6 decimal places.

For item A, $k=15$ and $t=3250$, so $N=15\lambda^{3250}$.

$N=15(0.9998784026\ldots)^{3250}=10.103\ldots$, so the current amount is $10.1$ g to 3 significant figures.

For item B, $18=25\lambda^t$, so $\lambda^t=\dfrac{18}{25}$.

$t\log\lambda=\log\left(\dfrac{18}{25}\right)$, so $t=\dfrac{\log(18/25)}{\log\lambda}$.

Using $\lambda=(1/2)^{1/5700}$ gives $t=2701\ldots$, so the age is $2700$ years to the nearest $100$ years.

The logarithms require $4x+5>0$ and $x+3>0$, so $x>-\dfrac54$.

$2\log_3(4x+5)-\log_3(x+3)=\log_3\left(\dfrac{(4x+5)^2}{x+3}\right)$.

$2=\log_3 9$, so $\dfrac{(4x+5)^2}{x+3}=9$.

$(4x+5)^2=9(x+3)$, so $16x^2+40x+25=9x+27$. Hence $16x^2+31x-2=0$.

$16x^2+31x-2=(16x-1)(x+2)=0$, so $x=\dfrac1{16}$ or $x=-2$.

$x=-2$ is outside the domain, so $x=\dfrac1{16}$.

$\log_{10}a+\log_{10}b=\log_{10}(a+b)$ gives $\log_{10}(ab)=\log_{10}(a+b)$.

Since the logarithm function is one-to-one, $ab=a+b$.

$ab-a=b$, so $a(b-1)=b$ and therefore $a=\dfrac{b}{b-1}$.

Given $a>0$ and $b>0$, the fraction $\dfrac{b}{b-1}$ must be positive. Since the numerator is positive, the denominator must also be positive, so $b>1$.

The logarithms require $4-x>0$ and $\dfrac{x+11}{2}>0$, so $-11<x<4$.

$2\log_2(4-x)=\log_2((4-x)^2)$ and $3+\log_2\left(\dfrac{x+11}{2}\right)=\log_2\left(8\cdot\dfrac{x+11}{2}\right)$.

Thus $(4-x)^2=4(x+11)$.

$x^2-8x+16=4x+44$, so $x^2-12x-28=0$. Hence $(x-14)(x+2)=0$.

The root $14$ is outside the domain, so $x=-2$.

At $P$, $3^{2x+1}=6\times3^x$.

Let $u=3^x$. Then $3^{2x+1}=3u^2$, so $3u^2=6u$.

Since $u>0$, divide by $3u$ to get $u=2$. Hence $3^x=2$ and $x=\log_3 2$.

$y=6\times3^x=6\times2=12$, so $P=(\log_3 2,12)$.

$3\log_3(2x-1)=\log_3(2x-1)^3$.

Since $2=\log_3 9$, the right side is $\log_3 9+\log_3(14x-25)=\log_3\{9(14x-25)\}$.

So $(2x-1)^3=9(14x-25)$.

$8x^3-12x^2+6x-1=126x-225$, which simplifies to $2x^3-3x^2-30x+56=0$.

Substituting $x=-4$ gives $2(-64)-3(16)-30(-4)+56=0$, so $-4$ is a root.

Dividing by $(x+4)$ gives $2x^2-11x+14=(2x-7)(x-2)$.

The possible roots are $-4$, $2$, and $\dfrac72$. The original logarithms require $2x-1>0$ and $14x-25>0$, so $x>\dfrac{25}{14}$.

Therefore the solutions are $x=2$ and $x=\dfrac72$.