Solve and check. 4^2x= 32^1/2 Can someone walk me through the steps? Thank you!! :)

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Solve and check. 4^2x= 32^1/2 Can someone walk me through the steps? Thank you!! :)

Algebra
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32 to the power of 1/2
@mthompson440 are you trying to simplify both sides of the equation?
I don't know..?

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32^1/2= 16
Divide both sides by 16.
\[\large4^{2x}=32^{1/2}\]\[\large (2^2)^{2x} = (2^5)^{1/2}\]\[\large 2^{4x} = 2^{5/2}\]\[\large 4x = \frac{5}{2}\]\[\large x=~?\]
Ok.. But I don't understand how to solve it.
@Jhannybean explained it
Your goal is to make the same base so you can evaluate their powers.
X= 5/8?
incorrect
32 = 2 * 2 * 2 * 2 * 2 4 = 2 * 2 Theres an exponent rule that states having the same base allows for you to evaluate powers. \[\large a^x = a^y \implies x=y\]
Yes @calecea . That is correct.
@Jhannybean i thought it was 1
@mthompson440 I kind of get it. I mean it's half of 32, but it was just confusing it's it's the exponent, not like multiplication.
\[4x=\frac{5}{2} \implies x = \frac{5}{2\cdot 4} = \frac{5}{8} \ne 1\]
@jhannybean thank you!
@mthompson440 32\(^{1/2}\) \(\ne\) 16
How do you get the 1/2 up there like that?
\[32^{1/2} =\sqrt{32} = \sqrt{16 \cdot 2} = \sqrt{4^2 \cdot 2} = 4\sqrt{2} \ne 16\]
what do you mean, "up there like that?"
like by the 32
\[32^1/2\]
`^{1/2}`
nvm figured it out
Have you learned exponential functions, @calecea ?
example problem: \(\large 7^{5x+3} = 512\)
Now try solving this one.
I have no idea. Probably.
Using our knowledge of log rules, we know that: \(\log (a)^b \implies b\log(a)\)
Okay, let's try that one instead.
\[\large \log(x) +\log(x+48)=2\]
Let me know how far you've gotten with it.

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