Question

wrong answers will be reported!!​

Answer 1

We're given to solve the equation,

$\longrightarrow\log_x2\times\log_{\frac{x}{16}}2=\log_{\frac{x}{64}}2$

Since $\log_ba=\dfrac{\log a}{\log b},$

$\longrightarrow\dfrac{\log2}{\log x}\times\dfrac{\log2}{\log\left(\dfrac{x}{16}\right)}=\dfrac{\log2}{\log\left(\dfrac{x}{64}\right)}$

Dividing by $\log2,$

$\longrightarrow\dfrac{\log2}{\log x}\times\dfrac{1}{\log\left(\dfrac{x}{16}\right)}=\dfrac{1}{\log\left(\dfrac{x}{64}\right)}$

$\longrightarrow\dfrac{\log2}{\log x\log\left(\dfrac{x}{16}\right)}=\dfrac{1}{\log\left(\dfrac{x}{64}\right)}$

Taking the reciprocal,

$\longrightarrow\dfrac{\log x\log\left(\dfrac{x}{16}\right)}{\log2}=\log\left(\dfrac{x}{64}\right)$

$\longrightarrow\log x\log\left(\dfrac{x}{16}\right)=\log2\log\left(\dfrac{x}{64}\right)$

Since $\log\left(\dfrac{a}{b}\right)=\log a-\log b,$

$\longrightarrow\log x\left(\log x-\log16\right)=\log2\left(\log x-\log64\right)$

Let $\log x=k.$ Then,

$\longrightarrow k\left(k-\log16\right)=\log2\left(k-\log64\right)$

$\longrightarrow k\left(k-\log(2^4)\right)=\log2\left(k-\log(2^6)\right)$

Since $\log(a^b)=b\,\log a,$

$\longrightarrow k\left(k-4\log2\right)=\log2\left(k-6\log2\right)$

$\longrightarrow k^2-4k\log2=k\log2-6\log^22$

$\longrightarrow k^2-5k\log2+6\log^22=0$

$\longrightarrow k^2-2k\log2-3k\log 2+6\log^22=0$

$\longrightarrow k(k-2\log2)-3\log 2(k-2\log2)=0$

$\longrightarrow(k-2\log2)(k-3\log2)=0$

$\Longrightarrow k\in\{2\log2,\ 3\log2\}$

$\longrightarrow\log x\in\{2\log2,\ 3\log2\}$

$\longrightarrow x\in\{2^2,\ 2^3\}$

$\longrightarrow\underline{\underline{x\in\{4,\ 8\}}}$

These are the solutions to the equation.

Answer 2

❍ Given :

${\red{\sf{\dag \: \: log_x \: 2 \: \times \: log_ \frac{x}{16} \: 2 \: = \: log_ \frac{x}{64} \: 2.}}}$

❍ Exigency To Do : We have to solve the following equation for x.

❍ Solution :

${\sf{\bigstar\:\:log_x \: 2 \: \times \: log_ \frac{x}{16} \: 2 \: = \: log_ \frac{x}{64} \: 2.}}$

We can write the equation as :

${\quad \longmapsto\: \: \sf{ \bigg( \dfrac{ \cancel{log \: 2}}{log \: x} \bigg) \bigg( \dfrac{log \: 2}{log \: \frac{x}{16} } \bigg) \: = \: \bigg( \dfrac{ \cancel{log \: 2}}{log \: \frac{x}{64} } \bigg) }}$

Here, log 2 is cancelled out, then we get :

$\quad \longmapsto \: \: \sf\frac{log \: 2}{log \: x \: (log \: x - 4 \: log \: 2 )} \: = \: \frac{1}{(log \: x - 6 \: log \: 2)}$

Then let us assume log x be t , then we get :

$\quad \longmapsto \: \: \sf\frac{log \: 2}{t(t - 4 \: log \: 2)} \: = \: \frac{1}{(t - 6 \: log \: 2)}$

Then by cross multiplication, we get :

$\quad \longmapsto \: \: \sf t \: log \: 2 - 6 \: (log \: 2) {}^{2} = t {}^{2} - 4 \: t \: log \: 2.$

$\quad \longmapsto \: \: \sf{t {}^{2} - 5 \: t \: log \: 2 + 6 \: (log \: 2) {}^{2} \: = \: 0}.$

Then value of t :

$\quad \sf \longmapsto \: \: t \: = \: 3 \: log \: 2 \: , \:2 \: log \: 2.$

$\quad \sf \longmapsto \: \: t \: = \: log \: 8 \: , \: log \: 4.$

Substituting the value of t :

$\quad \sf \longmapsto \: \: log\: x \: = \: log \: 8 \: , \:log \:x \:=\:log \: 4.$

Then value of x :

$\quad \purple{\sf {\dag \: \: x \: = \: 8 \: , \: x \: = \: 4 \:\:\checkmark}}.$

★ ═════════════════════ ★

❍ Answer :

$\quad \purple{\sf {\dag \: \: x \: = \: 8 \: , \: x \: = \: 4 \:\:\checkmark}}.$

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