# $x^{\log_b{a}}=2$ ; find $x$.

I was solving one problem in which I had to find the value of $$x$$ and at the last step my result came out to be $$x^{\log_b{a}}=2$$ but I was not able to get to the correct answer after this. Here's what I did:

\begin{align} &x^{\log_b{a}}=2\\ \Rightarrow\; & \log_b{a}\cdot\log_b{x}=\log_b{2} \rightarrow \text{Taking log to the base }b\text{ on both side}\\ \Rightarrow\; & \log_b{x}=\frac{\log_b{2}}{\log_b{a}}\\ \Rightarrow\; & \log_b{x}= \log_a{2} \end{align} But as you see I was not able to proceed ahead and I tried other ways to get to the solution but I was not able to.

The correct answer that has been provided is $$x=2^{\log_a{b}}$$. Can someone please help me as to how to get to this form of the answer?

• Note the edits I made to your question. In general, it is better to use double dollar signs to format multiline equations. The \begin{align}\end{align} environment helps to further typeset it so it looks nicer. I advise you review all my edits to see how they improve the post :)
– 5xum
Commented Apr 20, 2022 at 6:59
• Also, instead of writing $log$ for logarithms, which ends up looking like $log$ which looks like $l$, multiplied by $o$ multiplied by $g$, write $\log$ which typesets it as $\log$. Same for $sin,cos,arctan$ and so on which are all better as $\sin,\cos,\arctan$. And one final thing, use $\cdot$ for the multiplication sign instead of a period. $a\cdot b$ looks much nicer than $a.b$.
– 5xum
Commented Apr 20, 2022 at 7:01
• @5xum : Thanks! I will take care of these things from the next time. Commented Apr 20, 2022 at 7:35

In the reals, if $$\beta > 0$$, the solution to the equation $$x^{\alpha} = \beta$$ is $$x=\beta^\frac{1}{\alpha}$$ which is also sometimes written as $$x=\sqrt[\alpha]{\beta}$$.

Note that this is the same solution as the provided answer, because in your case, $$\beta=2$$ and $$\alpha=\log_b a$$, which means

$$x=2^{\frac{1}{\log_b a}} = 2^{\log_a b}$$

the last equality being true because, in general, $$\frac{1}{\log_b(a)} = \frac{1}{\frac{\ln a}{\ln b}} = \frac{\ln b}{\ln a} = \log_a(b)$$

Take $$\rm\log$$ w.r.t. $$\rm \, base\,2$$ so that $$\rm\log_ba \log_2x=1\implies \log_2x=\frac 1{\log_ba}=\log_ab\implies x=2^{\log_ba}$$

The first implication assumes $$\rm a\ne 1$$.

• $a\ne 1$ follows from $x^{\log_b a} = 2$ as $a= 1$ implies $1 = x^0 = \ne 2$ Commented Apr 20, 2022 at 7:21
• @martini: yes, I just wanted to emphasize and justify the division by $\log_ba$. :)
– Koro
Commented Apr 20, 2022 at 7:22