Evaluate log4 (81)

Question

Evaluate log4 (81)

zzr0ck3r · Answer

you sure that was not a 3?

anonymous · Answer

Positive.....it might be a mistake then

zzr0ck3r · Answer

there is not much you can do without a calculus class or a calculator

misty1212 · Answer

$$\frac{\ln(81)}{\ln(4)}$$

zzr0ck3r · Answer

that is not the same thing...

we want

$4^x=81$ you solved $x^4 = 81$

zzr0ck3r · Answer

@acxbox22

zzr0ck3r · Answer

But yes you could rewrite it as a ratio of different base logs, but that is not very "simplified" because you cant do anything with it...

acxbox22 · Answer

@zzr0ck3r oh yea sorry...

zzr0ck3r · Answer

no worries...

solomonzelman · Answer

$\LARGE\color{black}{   \log_4 (81) }$ 

$\LARGE\color{black}{   \log_4 (4^3) }$ 

$\LARGE\color{black}{   3\log_4 (4) }$   and on....

solomonzelman · Answer

there is a rule that  $\LARGE\color{black}{   \log_a (a)=1 }$   (when a>1)

zzr0ck3r · Answer

$4*4*4 \ne 81$

solomonzelman · Answer

Ohh my bad!

solomonzelman · Answer

it is 364 not 4^3, lol

solomonzelman · Answer

$\LARGE\color{black}{   \log_4 (3^4) }$ 

$\LARGE\color{black}{   4\log_4 (3) }$

zzr0ck3r · Answer

yeah, but not much nicer in my opinion.

zzr0ck3r · Answer

But I guess it depends on what you were doing with it:)

zzr0ck3r · Answer

its cool I just saw a question that asked them to solve ax+2b>8

and it did not tell them anything about a, it also was multiple choice so there were not separate cases answers available. ......

misty1212 · Answer

$$\huge \log_4(81)=\frac{\ln(81)}{\ln(4)}$$ by the change of base formula

anonymous · Answer

The change base formula solved the problem......Thanks guys