# Solve for x using properties of logarithms: log2(32)-log3(x)=log5(25)?

Feb 22, 2018

$x = 1$

#### Explanation:

Since ${2}^{5} = 3 \textcolor{w h i t e}{\text{xx")harrcolor(white)("xx}} {\log}_{2} \left(32\right) = 5$
and $\textcolor{w h i t e}{\text{x")5^2=25color(white)("xx")harrcolor(white)("xx}} {\log}_{5} \left(25\right) = 2$

${\log}_{2} \left(32\right) - {\log}_{3} \left(x\right) = {\log}_{5} \left(25\right)$
is equivalent to
$5 - {\log}_{3} \left(x\right) = 2$

$\Rightarrow {\log}_{3} \left(x\right) = 3$
and
since ${3}^{\textcolor{b l u e}{1}} = 3$
$\textcolor{w h i t e}{\text{XXX")log_3(x)=3color(white)("xx")rarrcolor(white)("xx}} x = \textcolor{b l u e}{1}$

Feb 22, 2018

$x = 27$

#### Explanation:

${\log}_{2} \left(32\right) - {\log}_{3} \left(x\right) = {\log}_{5} \left(25\right)$

${\log}_{2} \left({2}^{5}\right) - {\log}_{3} \left(x\right) = {\log}_{5} \left({5}^{2}\right)$

$5 {\log}_{2} \left(2\right) - {\log}_{3} \left(x\right) = 2 {\log}_{5} \left(5\right)$

$5 - {\log}_{3} \left(x\right) = 2$

${\log}_{3} \left(x\right) = 3$

Hence $x = {3}^{3} = 27$