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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. 
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. 

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algebra, logarithmic, logarithms, method, work 
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