My Math Forum Logarithmic algebra, why doesn't this method work?

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 September 4th, 2019, 05:38 AM #1 Newbie   Joined: Sep 2019 From: Sweden Posts: 1 Thanks: 0 Logarithmic algebra, why doesn't this method work? Problem is as follows: Solve the equation $\displaystyle 2^2x=5$ The book tells me to solve it by changing both sides to an exponentiation with the base 10. Logically, I tried to the following (Method 1): $\displaystyle (10^2)^2x=10^5$ $\displaystyle 10^4x=10^5$ $\displaystyle \log10^4x=\log10^5$ $\displaystyle 4x=5$ $\displaystyle x=5/4$ However, somehow, that isn't correct. The actual correct answer is $\displaystyle x=\log5/(2*\log2)$ The only hint the book gives me is $\displaystyle (10^(\log2))^2x=10^(\log5)$ I also tried the following (Method 2), and got the correct answer: $\displaystyle \log2^2x=\log5$ $\displaystyle 2x*\log2=\log5$ $\displaystyle 2x=\log5/\log2$ $\displaystyle x=\log5/(2*\log2)$ What I want to know is why Method 1 doesn't work, when Method 2 does. Last edited by skipjack; September 4th, 2019 at 10:16 AM.
 September 4th, 2019, 05:55 AM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,016 Thanks: 1600 $2^{2x} = 5$ $(10^{\log{2}})^{2x} = 10^{\log{5}}$ $10^{2x\log{2}} = 10^{\log{5}}$ $2x\log{2} = \log{5}$ $x = \dfrac{\log{5}}{2\log{2}}$ ... really don't know why they recommended this method when the standard method of taking the log of both sides at the beginning is much more efficient. Thanks from idontknow and EnitLee
 September 4th, 2019, 06:10 AM #3 Senior Member   Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 You claimed $\displaystyle 10^{(2^{2x})} = (10^2)^{2x}$. I'm pretty sure that is the error. ($\displaystyle x=5: 10^{1024} \neq 10^{40}$) By the way, use curly braces { } to group exponents and the like. Thanks from EnitLee

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