Re-arranging equations.

davehogan

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Sep 5, 2013
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I have to solve "3^x=5^(x-2)" using logarithms. I can follow this down to "2log5=xlog(5/3)". And then it finishes grandly with, "And finally x=2log5/log5/3". I don't understand how you get to that final answer from the previous equation. To get x to the left side you would need to divide both sides by x. Then, to get x on it's own you need to divide both sides by "2log5". This would give an entirely different answer. Can somebody explain this, please?
 

Subhotosh Khan

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I have to solve "3^x=5^(x-2)" using logarithms. I can follow this down to "2log5=xlog(5/3)". And then it finishes grandly with, "And finally x=2log5/log5/3". I don't understand how you get to that final answer from the previous equation. To get x to the left side you would need to divide both sides by x. Then, to get x on it's own you need to divide both sides by "2log5". This would give an entirely different answer. Can somebody explain this, please?
Certainly not!!

2log5=xlog(5/3)

(2log5)/x = log(5/3)

1/x = log(5/3)/(2log5)

x = (2log5)/[log(5/3)] .......................... same answer

better way

2log5=xlog(5/3)

(2log5)/[log(5/3)] = x

x = (2log5)/[log(5/3)].......................... same answer
 

JeffM

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Sep 14, 2012
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I have to solve "3^x=5^(x-2)" using logarithms. I can follow this down to "2log5=xlog(5/3)". And then it finishes grandly with, "And finally x=2log5/log5/3". I don't understand how you get to that final answer from the previous equation. To get x to the left side you would need to divide both sides by x. Then, to get x on it's own you need to divide both sides by "2log5". This would give an entirely different answer. Can somebody explain this, please?
\(\displaystyle 3^x = 5^{(x - 2)} \implies log(3^x) = log\left(5^{(x - 2)}\right).\)

Straightforward. If two expressions represent the same number, then the logs of the two expressions are equal because they they both represent the log of the same number.

\(\displaystyle log(3^x) = log\left(5^{(x - 2)}\right) \implies x * log(3) = (x - 2) * log(5).\)

Just applying one of the laws of logs to both sides of the equation. Now you seem to be getting rattled by logs. But they are just numbers.

\(\displaystyle Let\ u = log(5)\ and\ v = log(3).\)

\(\displaystyle x * log(3) = (x - 2) * log(5) \implies xv = (x - 2)u.\)

Simple substituion

\(\displaystyle xv = (x - 2)u = xu - 2u \implies xu - xv = 2u \implies x(u - v) = 2u \implies x = \dfrac{2u}{u - v}.\)

Just basic algebra.

\(\displaystyle x = \dfrac{2u}{u - v} \implies x = \dfrac{2log(5)}{log(5) - log(3)}.\)

Simple substitution.

\(\displaystyle x = \dfrac{2log(5)}{log(5) - log(3)} = \dfrac{2log(5)}{log\left(\dfrac{5}{3}\right)}\).

Another law of logs.
 

davehogan

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Sep 5, 2013
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Many thanks to both.
 
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