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log(base 10)3= a

log(base 10)2= b

Then log(base 5)6=???????

Please answer it.

10^a = 3

10^b = 2

(5^a)(2^a) = 3

(5^b)(2^b) = 2

5^b = 2^(1-b)

5^(b/(1-b)) = 2

(5^a)(5^(ab/(1-b))) = 3

5^(a/(1-b)) = 3

Multiply left sides and right sides of

5^(b/(1-b)) = 2

5^(a/(1-b)) = 3

To get

5^((a+b)/(1-b)) = 6

So log (base 5) 6 = ((a+b)/(1-b))

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I can answer any questions from the standard four semester Calulus sequence. Derivatives, partial derivatives, chain rule, single and multiple integrals, change of variable, sequences and series, vector integration (Green`s Theorem, Stokes, and Gauss) and applications. Pre-Calculus, Linear Algebra and Finite Math questions are also welcome.

Ph.D. in Mathematics and many years teaching undergraduate courses at three state universities. **Education/Credentials**

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