How to you answer the following question using logs:
What are the two solutions for x of 2^(2x)-8*2^x+15=0
In words: 2 to the power 2x, -8 multiplied by 2 to the power x, plus 15.
Please help!
$\displaystyle 2^{2x} - 8*2^x + 15 =(2^x)^2 - 8*2^x + 15=0$
$\displaystyle let \, u=2^x$
then this is just a quadratic equation in u, that is
$\displaystyle u^2 - 8u +15 = (u-3)(u-5)=0$
so u=3 or u=5
but since
$\displaystyle u= 2^x $
therefore $\displaystyle x=\frac{\ln 3}{\ln 2} = \log_2 3 \, or \, x=\frac{\ln 5}{\ln 2} = \log_2 5 $
From $\displaystyle 2^x=3\implies x=\log_23$ & $\displaystyle 2^x=5\implies x=\log_25.$
From this way, it's easier to check the solutions without a calculator.
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$\displaystyle 2^{2x}-8\cdot2^x+15=0.$
So for $\displaystyle x=\log_23,$
$\displaystyle 2^{2x}=2^{\log_29}=9$ & $\displaystyle 2^x=3,$ therefore
$\displaystyle 9-8\cdot3+15=0.$
You can do the same procedure to verify the second solution.