Why does $b^{\log_bx} = x$?

Question

Why does $b^{\log_bx} = x$?

Can someone break this down by showing me the steps as to why this is true?

Answer 1

It's the definition of the base-$b$ logarithm.

$\log_b c=d \Leftrightarrow b^d=c$.

So putting $y=\log_b x$ we have $b^y=x$. That is $b^{\log_b x}=x$.

Answer 2

$$b^{\log_b(x)}=b^{\frac{\ln(x)}{\ln(b)}}=\sqrt[\ln(b)]{b^{\ln(x)}}=\sqrt[\ln(b)]{x^{\ln(b)}}=x^{\frac{\ln(b)}{\ln(b)}}=x^1=x$$

Answer 3

Another way to think of this is to define a family of functions for any $b > 0$: $$f_b(x) = b^x.$$ Then there is an inverse function $$f_b^{-1}(x)$$ such that $$f_b^{-1}(f_b(x)) = x$$ for all real $x$, and $$f_b(f_b^{-1}(x)) = x$$ on the domain of the inverse function, which happens to be for $x > 0$. The inverse function is the logarithm: $$f_b^{-1}(x) = \log_b x.$$ Thus, $\log_b x^b = x$ and $b^{\log_b x} = x$ are consequences of the above and merely use different notation to express the same concept.

Another way to think about the base-$b$ logarithm is that logarithms are exponents. That is to say, $\log_b x$ is the exponent or power to which $b$ must be raised in order to obtain $x$.