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I'm doing a practice logarithm problem and I'm stuck. I'm given this reference

$$ \log_a2 = 0.3812\ \ \log_a3 = 0.6013\ \ \log_a5 = 0.9004 $$

And the question is

$$ \log_a(30a)^3 = x $$

The given answer choices are

$$ 3.169,\ 5.6487,\ 8.6487,\ 9.1435 $$

The way I try and solve it is

$$ \log_a2 = x \implies a^x = 2 \text{ and } a^x = y \implies a = y^{1/x} $$

So taking the first reference I should be able to find $a$ but the value I get from the three references is different. And I can't get a solution that matches the answer choices. What is it I'm missing?

Michael Hardy
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user193661
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2 Answers2

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Hint: $$ \log_a(30 a)^3= 3\log_a 30 +3 \log_a a=3(\log_a 2+\log_a3+\log_a5)+3 $$

Emilio Novati
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  • I thought the exponent on the outside of the parenthesis should not follow that property. – user193661 Jun 28 '17 at 19:45
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    why not? $\log_a(30 a)^3$ is usually interpreted as $ \log_a[(30a)^3]$ and it is different from $[\log(30a)]^3=\log^3(30a)$. – Emilio Novati Jun 28 '17 at 19:50
  • I mistakenly thought that the exponent on the outside meant that the log should be solved first then the exponent applied. – user193661 Jun 28 '17 at 19:54
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Let me suggest a different approach:

Following basic logarithm rules:

$$ \log_a(30a)^3 = 3(\log_a(2\times3\times5\times a)) = 3(\log_a(2)+\log_a(3)+\log_a(5)+\log_a(a)) $$

Can you proceed?

Dashi
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