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HI; I am struggling to why this works log base z to the power 3 (32) = log base 2 (32) / 3. Thanks.
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Idea said: \(\displaystyle \log _{a^m}\left(a^n\right)=\frac{n}{m}\) Click to expand... Nicely done.
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\(\displaystyle \log_{2^3} 32 =x\) \(\displaystyle (2^3)^x = 32\) \(\displaystyle 2^{3x} = 2^5\) 3x=5 x=5/3 \(\displaystyle \log_2 32 = 5\) Continue
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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 \(\displaystyle \log_{2^3}(32)=\dfrac{\log(32)}{\log(2^3)}=\dfrac{5\log(2)}{3\log(2)}\)
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\(\displaystyle \log _{a^m}\left(a^n\right)=\frac{n}{m}\)
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no the base is raised to the power 3 not the argument
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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
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I'm sorry yes there is it should be log base 2 to the power 3 (32) = log base 2 (32) / 3. Thanks
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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
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