Log Help ...

[Baron]

Simplify 2^(log (base 8) x^27)

So far I have
2^(log (base 2^3) x^(3*3))
2^(log (base 2) x^3)
2^(log x^3 / log 2)
2^(3log x / log2)

Then what do I do? I understand how to simplify logs but not when log is in the exponent.

 

[mmm4444bot]

Simplify 2^(log (base 8) x^27)

So far I have

Corrections shown in red

2^(log (base 2^3) x^(3*9))

2^(log (base 2) x^9)

At this point, we can stop.

By the definition of log_2(x^9), what happens when we raise 2 to that exponent ?

I mean, for example, log_2(8) = 3, yes?

So, 2^log_2(8) must equal 8 because that logarithm represents the exponent 3.

Got it?

 

[soroban]

Hello, Baron!

\(\displaystyle \text{Simplify: }\; 2^{\log_8(x^{27})}\) .[1]

\(\displaystyle \text{Let: }\log_8(x^{27}) \:=\\)

\(\displaystyle \text{Then: }8^P \:=\:x^{27} \quad\Rightarrow\quad (2^3)^P \:=\:x^{27} \quad\Rightarrow\quad 2^{3P} \:=\:x^{27}\)

\(\displaystyle \text{Take cube roots: }\:2^P \:=\:x^9\)

\(\displaystyle \text{And we have: }\ \:=\:\log_2(x^9)\)


Then [1] becomes:

. . \(\displaystyle 2^{\log_2(x^9)} \;=\; x^9\)