Logs

**KingV15** · January 19th 2010, 03:34 AM

Given that logab(a)=4, calculate logab 3√a /√b. Start with the exponent property of logs, then the quotient rule, and then change of base.

**Grandad** · January 19th 2010, 05:54 AM

Hello KingV15

Originally Posted by KingV15

Given that logab(a)=4, calculate logab 3√a /√b. Start with the exponent property of logs, then the quotient rule, and then change of base.

I'm guessing that the question is:

Given $\log_{ab}(a) = 4$ , calculate $\log_{ab}\left(\frac{\sqrt[3]{a}}{\sqrt{b}}\right)$

(The alternative is that you have to calculate $\log_{ab}\left(\frac{3\sqrt{a}}{\sqrt{b}}\right)$ , which will leave a term in $\log_{ab}(3)$ at the end.)

So, if my assumption is correct:

$\log_{ab}(a) = 4$

$\Rightarrow (ab)^4 = a$

$\Rightarrow a^4b^4 = a$

$\Rightarrow b^4 = a^{-3}$

$\Rightarrow b = a ^{-\frac34}$

And then:

$\log_{ab}\left(\frac{\sqrt[3]{a}}{\sqrt{b}}\right)=\tfrac13\log_{ab}(a) - \tfrac12\log_{ab}(b)$

$=\tfrac13\log_{ab}(a) - \tfrac12\log_{ab}(a^{-\frac34})$

$=\tfrac13\log_{ab}(a) - \tfrac12\times(-\tfrac34)\log_{ab}(a)$

$=(\tfrac13+\tfrac38)\log_{ab}(a)$

$=\frac{17}{24}\times 4$

$=\frac{17}{6}$

(The answer to the alternative question is, using a similar method, $\frac72+\log_{ab}(3)$ .)
Grandad

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