1
$\begingroup$

I am reviewing algebra 2 using some video tutorial from mathtutordvd.com.

In one of the videos, the author converts a logarithm equation into the exponent form as follows:

enter image description here

Which logarithm identity is used to convert the equation?

$\endgroup$
  • $\begingroup$ That's just because the exponential function is well defined. $\endgroup$ – mwt Sep 26 '18 at 18:58
  • $\begingroup$ They just took the exponential of both sides... if $x=y$ then $6^x=6^y$ $\endgroup$ – user438666 Sep 26 '18 at 19:03
  • $\begingroup$ This being said, it is far too complicated. Logarithms in any base are bijections from $\mathbf R_+$ to $\mathbf R$, so $\log_n(2x-3)=\log_b 4$ implies $2x-3=4$. $\endgroup$ – Bernard Sep 26 '18 at 19:03
1
$\begingroup$

We are using the definition

$$\log_a b=c\iff a^c=a^{\log_a b}=b$$

but since $\log$ function is injective we can conclude directly

$$\log_a f(x)=\log_a g(x)\iff f(x)=g(x)$$

for $f(x),g(x)>0$.

$\endgroup$
  • $\begingroup$ Got it. Thanks! $\endgroup$ – Cody Sep 26 '18 at 19:03
  • $\begingroup$ You are welcome! Bye $\endgroup$ – user Sep 26 '18 at 19:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.