Question

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

Answer 1

`x=1/3`

Explanation:

Log form and index form are interchangeable.

`log_a b = c " "hArr" " a^c =b`

Apply that to the equation given:

`log_10 root3(10) =x" "hArr" " 10^x = root3(10)`

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

`10^x= 10^(1/3)`

`:. x =1/3`

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

`(10^x)^3 = (root3(10))^3`

`10^(3x) = 10^1`

`3x =1`

`x =1/3`

Answer 2

`x=1/3=0.bar3`

Explanation:

Given: `log_10root(3)(10)=x`.

A basic property of logs is that:

If `log_ab=m` then `a^m=b`.

So, we get:

`10^x=root(3)10`

Note that `root(3)10=10^(1/3)` due to the fact that `root(a)(m^n)=m^(n/a)`.

Therefore,

`10^x=10^(1/3)`

It is now clear to see that `x=1/3`. In decimal form, that is `0.3333...=0.bar3`.