Number raised to log expression I am struggling with what I think should be some a basic log problem: Show that 3^(log_2n)=n^(log_23) I know that 3^(log_3n)=n and log2n=log3n/log32 I was attempting something similar to: 3^(log_3n/log_32)=3^(log_3n−log_32) but then I got stuck. Am I on the right track by using the change of base and then subtracting? EDIT: I'm not trying to exactly 'show that' the two expressions are equal. I am looking to figure out how to simplify the expression on the left to the one on the right

$\frac{20b}{{\left(4b^3\right)}^3}$

Which operation could we perform in order to find the number of milliseconds in a year??
${\displaystyle 60\cdot 60\cdot 24\cdot 7\cdot 365}$
${\displaystyle 1000\cdot 60\cdot 60\cdot 24\cdot 365}$
${\displaystyle 24\cdot 60\cdot 100\cdot 7\cdot 52}$
${\displaystyle 1000\cdot 60\cdot 24\cdot 7\cdot 52?}$

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A) No
B) 0
C) Yes
D) 6

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149600000000 is equal to
A)$1.496\times <!--\; \times \; -->{10}^{11}$
B)$1.496\times <!--\; \times \; -->{10}^{10}$
C)$1.496\times <!--\; \times \; -->{10}^{12}$
D)$1.496\times <!--\; \times \; -->{10}^8$

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