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 February 10th, 2017, 07:29 AM #1 Newbie   Joined: Oct 2014 From: In the sea Posts: 15 Thanks: 1 Logarithm function Hi, If*$\displaystyle log10(b)=1.8*and*loga(b)=2.5752$, what is*a? How does the change of base rule apply here when there are two unknowns a & B ? Thanks, amphi
February 10th, 2017, 07:54 AM   #2
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Quote:
 Originally Posted by amphinomos Hi, If*$\displaystyle log10(b)=1.8*and*loga(b)=2.5752$, what is*a? How does the change of base rule apply here when there are two unknowns a & B ? Thanks, amphi
$\log_a{b} = \dfrac{\log_{10}{b}}{\log_{10}{a}}$

$2.5752 = \dfrac{1.8}{\log_{10}{a}}$

$\log_{10}{a}= \dfrac{1.8}{2.5752}$

$a = 10^{\frac{1.8}{2.5752}}$

 February 27th, 2017, 01:14 PM #3 Newbie   Joined: Feb 2017 From: Omaha, Nebraska Posts: 8 Thanks: 0 Math Focus: Algebra and Trigonometry Loagarithms equate to Exponents Really like Skeeter's solution, but when working with logarithms I get to exponents as soon as possible... log b = 1.8 loga b = 2.5752 logb a = 1/2.5752 b^(1/2.5752) = a 10^(1.8 (1/2.5752)) = a 10^(1.8/2.5752) = a Since logarithms equate to exponents. Last edited by y2kevin; February 27th, 2017 at 01:23 PM. Reason: clearify math.

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