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September 4th, 2019, 05:38 AM   #1
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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.
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September 4th, 2019, 05:55 AM   #2
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$2^{2x} = 5$

$(10^{\log{2}})^{2x} = 10^{\log{5}}$

$10^{2x\log{2}} = 10^{\log{5}}$

$2x\log{2} = \log{5}$

$x = \dfrac{\log{5}}{2\log{2}}$

... 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.
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September 4th, 2019, 06:10 AM   #3
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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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