Help solve 2^(1-2x)=3?
I'm learning of logs and functions :) I'm not quite sure where to go about solving this though :c
It looks like this: \[2^{(1-2x)=3}\] What you need to do, is solve the exponent like an equation, then substitute in, and solve.
=3 isn't in the exponent, is that alright?
@ParthKohli Help D:
Your main goal is to get the \(1 - 2x\) out of the exponent so you're able to solve the equation. Do you know that you can take the log of both the sides? Try doing that and see what you get.
Let's solve the exponent: \[(1-2x)=3\] \[2x = 4\] \[x = 4\] Now substitute in. \[(1-2[4])=3\] \[(1-8)=3\] -7 = 3 Now... \[2^{-7}\] Can you solve from there?
I'm afraid that's not correct, @AnyTipzWouldHelp
I messed up, sorry.
it has a log base of 2
Am I supposed to have (1-2x) log 2 = log 3
Yes, very good. Now get the \(x\) alone.
The first step would be to divide both the sides by \(\log(2)\)
So (1-2x)=log3/log2?
Or no?
@ParthKohli
Yes. Or \(1 - 2 x = \log_2 3\).
Then what :)
Just "isolate" \(x\). For example, subtract 1 from both sides.
My answer options are : x = 0.396 x = -0.396 x = -0.631 x = 0.631 Just to make sure we're going down the right track
3/2-1? This would equal .5
We very much are.
No, that will become \(-2x = \log_2(3) - 1\).
Alright, so don't literally solve it, but show the subtraction
Yes. What's the value of \(\log_{2}3\) according to your calculator?
How should I enter it? Just as log(2)3?
When I do that, I get .903089987
Have you heard the base-change property?
heard about*
I'm familiar with it, I could use a refresher
It just says that\[\log_a b = \dfrac{\log b}{\log a}\]
alright, I got 1.584962501 :)
Nice. Now just solve for \(x\).
Do I need to subtract one from this first? And then divide by negative two?
I got -.2924812505, is this correct @ParthKohli
Seems good. :)
Doesn't seem to be one of the choices. I wonder why.
LOL! I see that I accidentally wrote the wrong ones.. I chose answer choice -.292 :)
Right.
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