Logarithms: CAT previous year questions

Solution

The base $\dfrac{1}{4}$ is less than 1, so $\log_{1/4}(t) > 0$ means $0 < t < 1$.

We need $0 < n^2 - 7n + 11 < 1$. The upper bound gives $n^2 - 7n + 10 < 0 \Rightarrow 2 < n < 5$, so $n = 3$ or $4$.

But $n = 3$ gives $-1$ and $n = 4$ gives $-1$, which are not positive.

No integer works: $0$.

Solution

Read the equation as $\log_{64} x^2 + \log_8 \sqrt{y} + 3\log_{512}(\sqrt{y}\,z) = 4$ and convert to base 2:

$\dfrac{2}{6}\log_2 x + \dfrac{1}{6}\log_2 y + \dfrac{3}{9}\left(\dfrac{1}{2}\log_2 y + \log_2 z\right) = 4$.

$\dfrac{1}{3}(\log_2 x + \log_2 y + \log_2 z) = 4 \Rightarrow xyz = 2^{12} = 4096$.

By AM–GM, $x + y + z \geq 3\sqrt[3]{4096} = 3 \times 16 = 48$.

Solution

The base $x - 3$ must be positive and not 1: $x > 3$, $x \neq 4$.

$\log_{x-3}(x^2 - 9) = \log_{x-3}\left[(x + 1)(x - 3)^2\right]$, so $(x - 3)(x + 3) = (x + 1)(x - 3)^2$.

$x + 3 = (x + 1)(x - 3) \Rightarrow x^2 - 3x - 6 = 0 \Rightarrow x = \dfrac{3 \pm \sqrt{33}}{2}$.

Only $\dfrac{3 + \sqrt{33}}{2} \approx 4.37$ satisfies $x > 3$. Sum $= \dfrac{3 + \sqrt{33}}{2}$.

Solution

Let $L = \log_{10} x$. Then $\log_{100} x = \dfrac{L}{2}$ and $\log_{1000} x = \dfrac{L}{3}$.

$4L + 2L + \dfrac{8L}{3} = \dfrac{26L}{3} = 13 \Rightarrow L = 1.5$.

$x = 10^{1.5} = 10\sqrt{10} \approx 31.6$. Greatest integer $= 31$.

Solution

$\dfrac{\log_8(a + b)}{\log_2 c} = \dfrac{1}{3}\log_c(a + b)$ and $\dfrac{\log_{27}(a - b)}{\log_3 c} = \dfrac{1}{3}\log_c(a - b)$.

So $\dfrac{1}{3}\log_c(a^2 - b^2) = \dfrac{2}{3} \Rightarrow a^2 - b^2 = c^2 \Rightarrow a^2 = b^2 + c^2$.

With $b, c \leq 10$: $a^2 \leq 200 \Rightarrow a \leq 14.1$.

$a = 14$ is possible (e.g. $b = c = \sqrt{98} \approx 9.9$). Greatest integer $a = 14$.

Solution

Let $t = 10^x$: $t + \dfrac{4}{t} = \dfrac{81}{2} \Rightarrow 2t^2 - 81t + 8 = 0$, with two positive roots.

Product of roots $t_1 t_2 = 4 \Rightarrow 10^{x_1 + x_2} = 4$.

$x_1 + x_2 = \log_{10} 4 = 2\log_{10} 2$.

Solution

$a = \log_3 4$, $b = \log_4 5$, …, $f = \log_8 9$.

The chain telescopes: $abcdef = \log_3 9 = 2$.

Solution

$\log_x(x^2 + 12) = 4 \Rightarrow x^4 = x^2 + 12 \Rightarrow x^2 = 4 \Rightarrow x = 2$.

$3\log_y 2 = 1 \Rightarrow \log_y 2 = \dfrac{1}{3} \Rightarrow y = 8$.

$x + y = 10$.

Solution

$\dfrac{\log_x 25}{\log_x 0.008} = \log_{0.008} 25 = \dfrac{\log 5^2}{\log 5^{-3}} = -\dfrac{2}{3}$.

So $\log_{\sqrt{3}} x = \dfrac{16}{3} + \dfrac{2}{3} = 6 \Rightarrow x = (\sqrt{3})^6 = 27$.

$\log_3(3x^2) = 1 + 2\log_3 27 = 1 + 6 = 7$.

Solution

The terms are $\dfrac{1}{2}$, $\log_4(2^x - 9)$ and $\log_4\left(2^x + \dfrac{17}{2}\right)$.

AP: $2\log_4(t - 9) = \dfrac{1}{2} + \log_4\left(t + \dfrac{17}{2}\right)$ with $t = 2^x$, i.e. $(t - 9)^2 = 2\left(t + \dfrac{17}{2}\right) = 2t + 17$.

$t^2 - 20t + 64 = 0 \Rightarrow t = 16$ (as $t > 9$).

Common difference $= \log_4 7 - \dfrac{1}{2} = \log_4 7 - \log_4 2 = \log_4\dfrac{7}{2}$.

Solution

Let $L = \log_2 n$; then $\log_4 n = \dfrac{L}{2}$.

$\dfrac{4 - L}{3 - L/2} = \dfrac{2(4 - L)}{6 - L} < 0 \Rightarrow 4 < L < 6 \Rightarrow 16 < n < 64$.

$n = 17, \dots, 63$: $47$ values.

Solution

Write all terms with $\log = \log_{10}$:

$5 - \dfrac{1}{2}\log(1 + x) + 2\log(1 - x) = -\dfrac{1}{2}\log(1 - x) - \dfrac{1}{2}\log(1 + x)$.

$5 + \dfrac{5}{2}\log(1 - x) = 0 \Rightarrow \log(1 - x) = -2 \Rightarrow 1 - x = 0.01$.

$x = 0.99$, so $100x = 99$.

Solution

$\log_2[\dots] = 2 \Rightarrow 3 + \log_3\{\dots\} = 4 \Rightarrow \log_3\{4 + \log_4(x - 1)\} = 1$.

$4 + \log_4(x - 1) = 3 \Rightarrow \log_4(x - 1) = -1 \Rightarrow x - 1 = \dfrac{1}{4}$.

$x = \dfrac{5}{4}$, so $4x = 5$.

Solution

$\dfrac{p + q}{pq} = \dfrac{1}{q} + \dfrac{1}{p}$ with $p = \log_{15} a$, $q = \log_{32} a$.

$= \log_a 32 + \log_a 15 = \log_a 480 = 4 \Rightarrow a^4 = 480$.

$4^4 = 256 < 480 < 625 = 5^4$, so $4 < a < 5$.

Solution

Take $\log_2$ of both sides: $Y^2 \log_3 5 = \log_2 3 \cdot \log_2 5$.

$\dfrac{\log_2 5}{\log_3 5} = \log_2 3$, so $Y^2 = (\log_2 3)^2$.

$Y$ is negative: $Y = -\log_2 3 = \log_2\dfrac{1}{3}$.

Solution

$\log_6\sqrt{5} = \dfrac{1}{2}\log_6 5$. Using natural logs:

$\dfrac{\ln 5}{\ln 4} = \dfrac{\ln y}{\ln 4} \cdot \dfrac{\ln 5}{2\ln 6} \Rightarrow \ln y = 2\ln 6$.

$y = 36$.

Solution

$\log_a\dfrac{a}{b} + \log_b\dfrac{b}{a} = 2 - \left(t + \dfrac{1}{t}\right)$, where $t = \log_a b \geq 1$.

$t + \dfrac{1}{t} \geq 2$, so the value is at most 0.

It cannot equal $1$.

Solution

$\log_2 a = \dfrac{1}{3} \Rightarrow \log_a 2 = 3$.

$A = \log_a 30 = 3 + \log_a 3 + \log_a 5$ and $-B = \log_a 5 - \log_a 3$.

Subtracting: $A + B - 3 = 2\log_a 3 \Rightarrow \log_a 3 = \dfrac{A + B - 3}{2}$.

$\log_3 a = \dfrac{2}{A + B - 3}$.

Solution

Let $14^a = 36^b = 84^c = k$. Then $\dfrac{1}{a} = \dfrac{\ln 14}{\ln k}$, $\dfrac{1}{c} = \dfrac{\ln 84}{\ln k}$, $b = \dfrac{\ln k}{\ln 36}$.

$\dfrac{1}{c} - \dfrac{1}{a} = \dfrac{\ln 6}{\ln k}$ and $6b = \dfrac{6\ln k}{2\ln 6} = \dfrac{3\ln k}{\ln 6}$.

Product $= 3$.

Solution

Numerator $= 2 \times 4 \times 8 \times 16 = 2^{10} = 1024$.

$\log_2 4 = 2$, $\log_4 8 = \dfrac{3}{2}$, $\log_8 16 = \dfrac{4}{3}$.

Denominator $= 2^2 \times \left(\dfrac{3}{2}\right)^3 \times \left(\dfrac{4}{3}\right)^4 = 4 \times \dfrac{27}{8} \times \dfrac{256}{81} = \dfrac{128}{3}$.

Value $= 1024 \times \dfrac{3}{128} = 24$.