# If log_ba=1/x and log_a √b =3x^2, show that x =1/6?

Jun 3, 2018

$\text{see explanation}$

#### Explanation:

$\text{using the "color(blue)"law of logarithms}$

•color(white)(x)log_b x=nhArrx=b^n

${\log}_{b} a = \frac{1}{x} \Rightarrow a = {b}^{\frac{1}{x}}$

${\log}_{a} {b}^{\frac{1}{2}} = 3 {x}^{2} \Rightarrow {b}^{\frac{1}{2}} = {a}^{3 {x}^{2}}$

$\text{substitute "a=b^(1/x)" into } {a}^{3 {x}^{2}}$

${b}^{\frac{1}{2}} = {\left({b}^{\frac{1}{x}}\right)}^{3 {x}^{2}} = {b}^{3 x}$

$\text{we have } {b}^{\frac{1}{2}} = {b}^{3 x}$

$3 x = \frac{1}{2}$

$\text{divide both sides by 3}$

$x = \frac{\frac{1}{2}}{3} = \frac{1}{6}$

Jun 3, 2018

$x = \frac{1}{6}$

#### Explanation:

${\log}_{b} a = \frac{1}{x}$

So

$x {\log}_{b} a = 1$

$x = \frac{1}{\log} _ b a$

$x = {\log}_{a} b$

Now

$\frac{1}{2} {\log}_{a} b = 3 {x}^{2} \text{ }$ (substituting from ${\log}_{a} b = x$)

So

$\frac{x}{2} = 3 {x}^{2}$

$x = 6 {x}^{2}$

$6 x = 1$

$x = \frac{1}{6}$ with $x \ne 0$

Hence proved.

Jun 3, 2018

We will start the proof

#### Explanation:

We have
$\ln \frac{a}{\ln} \left(b\right) = \frac{1}{x}$
and

$\frac{1}{2} \cdot \ln \frac{b}{\ln} \left(a\right) = 3 {x}^{2}$
so

$\ln \frac{a}{\ln} \left(b\right) = \frac{1}{6 {x}^{2}}$

using the first equation we get
$\frac{1}{x} = \frac{1}{6 {x}^{2}}$

multiplying by ${x}^{2}$ we get

$x = \frac{1}{6}$ for $x \ne 0$