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?
Re-arranging equations.
- Thread starter davehogan
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Certainly not!!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?
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
\(\displaystyle 3^x = 5^{(x - 2)} \implies log(3^x) = log\left(5^{(x - 2)}\right).\)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?
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.