Question

The value of $x$ satisfying the equation ${\left[3(1-\frac{1}{2}+\frac{1}{4}....\infty )\right]}^{{\log }_{10}x}={\left[20(1-\frac{1}{4}+\frac{1}{16}....\infty )\right]}^{{\log }_{x}10}$ is

• A. $\frac{1}{100}$
• B. $10$
• C. $1000$
• D. $10000$

Answer 1

Answer: (A)

$\frac{1}{100}$

$1-\frac{1}{2}+\frac{1}{4}+...\infty$

$=\frac{1}{1-\left(\frac{-1}{2}\right)}$... since the common ratio is -0.5

$=\frac{2}{2+1}$

$=\frac{2}{3}$.

Hence

$3(1-\frac{1}{2}+\frac{1}{4}+...\infty )$

$=3\left(\frac{2}{3}\right)$

$=2$ ...(i)

Similarly

$20(1-\frac{1}{4}+\frac{1}{16}+...\infty )$

$=20\left(\frac{1}{1-\left(\frac{-1}{4}\right)}\right)$

$=20\left(\frac{4}{4+1}\right)$

$=20\left(\frac{4}{5}\right)$

$=4\times 4$
$=16$ ...(ii)

Hence

$2^{lo{g}_{10}\left(x\right)}={16}^{lo{g}_x\left(10\right)}$

$2^{lo{g}_{10}\left(x\right)}=2^{4lo{g}_x\left(10\right)}$

$2^{lo{g}_{10}\left(x\right)}=2^{lo{g}_x{10}^4}$

Comparing the exponents give us

$lo{g}_{10}\left(x\right)=4lo{g}_x\left(10\right)$

$\frac{log10}{logx}=\frac{logx}{4log10}$

$4lo{g}^{2}10=lo{g}^{2}x$

$log\left(x\right)=\pm 2log\left(10\right)$

$x=100$ and $x=\frac{1}{100}$

Hence there are two solutions

$x=\frac{1}{100},100$.