logarithms in different form

Question

HI; I am struggling to why this works log base z to the power 3 (32) = log base 2 (32) / 3. Thanks.

Answer 1

Idea said: \(\displaystyle \log _{a^m}\left(a^n\right)=\frac{n}{m}\) Click to expand... Nicely done.

Answer 2

\(\displaystyle \log_{2^3} 32 =x\) \(\displaystyle (2^3)^x = 32\) \(\displaystyle 2^{3x} = 2^5\) 3x=5 x=5/3 \(\displaystyle \log_2 32 = 5\) Continue

Answer 3

anthonye said: log base 2 to the power 3 (32) = log base 2 (32) / 3. Click to expand... The standard change of base is \(\displaystyle \log_b(a)=\dfrac{\log(a)}{\log(b)}\). So \(\displaystyle \log_{2^3}(32)=\dfrac{\log(32)}{\log(2^3)}=\dfrac{5\log(2)}{3\log(2)}\)

Answer 4

\(\displaystyle \log _{a^m}\left(a^n\right)=\frac{n}{m}\)

Answer 5

no the base is raised to the power 3 not the argument

Answer 6

anthonye said: I'm sorry yes there is it should be log base 2 to the power 3 (32) = log base 2 (32) / 3. Thanks Click to expand... Okay, so this is supposed to be \(\displaystyle log_2 (32 ^3) = \dfrac{1}{3} log_2 (32)\)? If so then that isn't correct. \(\displaystyle log_2 (32 ^3) = 3 ~ log_2 (32)\). -Dan

Answer 7

I'm sorry yes there is it should be log base 2 to the power 3 (32) = log base 2 (32) / 3. Thanks

Answer 8

anthonye said: HI; I am struggling to why this works log base z to the power 3 (32) = log base 2 (32) / 3. Thanks. Click to expand... \(\displaystyle log_z (32^3)\) cannot be decomposed into anything without a z in it. Is there a typo? You are probably looking at a change of base formula problem: \(\displaystyle log_a b = \dfrac{log_c b}{log_c a}\) -Dan