<p>The explanation in Barron's is terrible. </p>
<p>If log(8) 3 = x*log(2) 3 then x = </p>
<p>a) 1/3
b) 3
c) 4
d) log(4) 3
e) log(8) 9</p>
<p>the answer is a but how do you get there? </p>
<p>thanks</p>
<p>x = log(8)3/log(2)3. </p>
<p>log(8)3 = log (2^3)3=1/3*log(2)3 => x=1/3</p>
<p>Use change of base – log<em>8 3 = (log</em>2 3)/(log<em>2 8) = (log</em>2 3)/3.</p>
<p>so you just have to memorize the change of base formula?</p>
<p>Yes, but not too difficult to memorize. Just remember that the 3 (or the number you are trying to take the logarithm of) goes on top, and the old base goes on bottom.</p>
<p>These would also be helpful to memorize:
log<em>b^n a^m = (m/n) log</em>b a
log<em>b a^m = m log</em>b a
log<em>b^n a = (1/n) log</em>b a.</p>
<p>log<em>8 3 = log</em>2^3 3 = (1/3) log<em>2 3 = x log</em>2 3
x = 1/3</p>
<p>^ This is pretty much what minnan did, just with log identities.</p>
<p>@Posterguy: yes, you should memorize the formula as gcf101 recited.</p>