# How do you solve log_10 root3(10)=x?

Jul 9, 2018

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

#### Explanation:

Log form and index form are interchangeable.

${\log}_{a} b = c \text{ "hArr" } {a}^{c} = b$

Apply that to the equation given:

${\log}_{10} \sqrt{10} = x \text{ "hArr" } {10}^{x} = \sqrt{10}$

To solve an exponential equation, one approach is to make the bases the same:

${10}^{x} = {10}^{\frac{1}{3}}$

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

We could also have raised the whole equation to the power of $3$

${\left({10}^{x}\right)}^{3} = {\left(\sqrt{10}\right)}^{3}$

${10}^{3 x} = {10}^{1}$

$3 x = 1$

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

Jul 9, 2018

$x = \frac{1}{3} = 0. \overline{3}$

#### Explanation:

Given: ${\log}_{10} \sqrt{10} = x$.

A basic property of logs is that:

If ${\log}_{a} b = m$ then ${a}^{m} = b$.

So, we get:

${10}^{x} = \sqrt{10}$

Note that $\sqrt{10} = {10}^{\frac{1}{3}}$ due to the fact that $\sqrt[a]{{m}^{n}} = {m}^{\frac{n}{a}}$.

Therefore,

${10}^{x} = {10}^{\frac{1}{3}}$

It is now clear to see that $x = \frac{1}{3}$. In decimal form, that is $0.3333 \ldots = 0. \overline{3}$.