Playing with log

Algebra Level 4

If ${a^{\log_a x}=x^{2}}$ where $a$ is any positive real number.

Then, evaluate ${ e^{x}}$

1 & e 1

\ln x

\frac{1} {e}

Indeterminate 0

\ln (1-x)^{2}

1 solution

Md Omur Faruque
Jul 13, 2015

Applying $a^{log_a x}=x$ , we get, $x=x^{2}$ $x^{2}-x=0$ $x(x-1)=0$ $x=0,1$

But, when x=0, $\log_a 0$ is undefined.

So, the answer will only be $\color{#69047E} {\boxed{e^{1}=e}}$ .

Moderator note:

When applying a formula, we have to check what are the conditions under which it is valid. For example, the stated formula of

$a ^ { \log a x } = x$

is only true if $a > 0, x > 0$ . This is why we had to reject $x = 0$ .

When manipulating an equation, we always have to be careful of introducing extraneous roots. It's important to check that the solutions that we get are indeed valid.

I haven't thought that!!! it's quite confusing!

Jayed Taher - 5 years, 11 months ago

@Challenge Master Note

Sir, the question was $a^{log_a x}=x$ , not $a^{\log ax}=x$

MD Omur Faruque - 5 years, 10 months ago

Playing with log

1 solution

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