Question

Solve for x :
$lo{g}_{3/4}lo{g}_8(x^2+7)+lo{g}_{1/2}lo{g}_{1/4}(x^2+7{)}^{-1}=-2$

• A. $x=3$
• B. $x=5$
• C. $x=-3$
• D. $x=6$

Answer 1

Answer: (A), (C)

The correct options are
A $x=3$
C $x=-3$

$lo{g}_{3/4}\left[\frac{1}{3}lo{g}_2\right(x^2+7\left)\right]+lo{g}_{1/2}\left[\frac{1}{2}lo{g}_2\right(x^2+7\left)\right]=-2$

By $lo{g}_{b^p}{a}^n=\frac{n}{p}lo{g}_{b}a$

$lo{g}_{2^{-2}}(x^2+7{)}^{-1}=\frac{-1}{-1}lo{g}_2(x^2+7)$

Let $lo{g}_2(x^2+7)=t\therefore x^2+7=2^t$

$lo{g}_{3/4}\left(\frac{1}{3}t\right)+lo{g}_{1/2}\left(\frac{1}{2}t\right)=-2$

$\frac{log\left(t/3\right)}{log\left(3/4\right)}+lo{g}_{1/2}\frac{1}{2}+lo{g}_{1/2}t=-2$

$\frac{logt-log3}{log3-log4}+lo{g}_{1/2}\frac{1}{2}+lo{g}_{1/2}t=-2$

$\frac{logt}{log3-2log2}-\frac{log3}{log3-2log2}+1-\frac{logt}{log2}=-2$

$\therefore logt\frac{[log2-log3+2log2]}{(log3-2log2)log2}$

$=-3+\frac{log3}{log3-2log2}$

$logt\frac{[3log2-log3]}{(log3-2log2)log2}=\frac{6log2-2log3}{log3-2log2}$

$logt=2log2=log{2}^2=log4$

$\therefore t=4\therefore x^2+7=2^t=2^4=16$

or $x^2=9\therefore x=\pm 3$