Question

Find the value of x if $(2x{)}^{ln2}$ = $(3y{)}^{ln3}$ and $3^{lnx}$ = $2^{lny}$.

${\log }_{e}x$ can also be written as ln(x)

__

Answer 1

$(2x{)}^{ln2}$ = $(3y{)}^{ln3}$ ------------(1)

$3^{lnx}$ = $2^{lny}$ -------------------(2)

Taking log on both sides of equation 1

ln2 ln(2x) = ln3 (ln(3y))

ln2 $\times$ [ln2 + lnx] = ln3 $\times$ [ln3 + lny]

$(ln2{)}^2$ + ln2 $\times$ lnx = $(ln3{)}^2$ + ln3 $\times$ lny----------(3)

Taking log on both sides of equation 2

lnx ln3 = lny ln2 $\Rightarrow$ lny = $\frac{lnx.ln3}{ln2}$ ----------(4)

Substitute lny value from equation 4 to equation 3

$(ln2{)}^2$ + ln2 $\times$ lnx = $(ln3{)}^2$ + ln3 $\times$ $\frac{lnx.ln3}{ln2}$

$(ln2{)}^2$ + ln2 $\times$ lnx = $(ln3{)}^2$ + $(ln3{)}^2$ . $\frac{lnx}{ln2}$

$(ln2{)}^2$ - $(ln3{)}^2$ = $(ln3{)}^2$ . $\frac{lnx}{ln2}$ - ln2 . Inx
$(ln2{)}^2-(ln3{)}^2=\frac{lnx}{ln2}\left[\right(ln3{)}^2-\left(ln2{)}^2\right]$
$\Rightarrow lnx=-ln2$
$\Rightarrow x=\frac{1}{2}=0.5$