Solve 4^(2x) = 7^(x−1).
@Astrophysics @rishavraj @ganeshie8
have you tried taking ln( ) of both sides
wouldn't it be log?
\[\ln(4^{2x})=\ln(7^{x-1}) \\ \\ \text{ Now use power rule } \\ 2x \ln(4)=(x-1) \ln(7)\] you can use log too if you want
not distribute on the right hand side
and put your terms with x in it on one side and your terms without x on the opposing side
1.38629436112(2x) I mean, sorry :)
are you allowed to approximate ?
no, not until we have the final answer
ok so leave that one side as 2ln(4)x and distribute on the other side recall the distributive property is a(b+c)=ab+ac
use distributive property here and what do you get: \[\ln(7)(x-1)=?\]
1.94591014906
1.94591014906x-1.94591014906
I think you mean to say ln(7)x-ln(7)
oh yes! Sorry
\[\ln(4^{2x})=\ln(7^{x-1}) \\ \\ \text{ Now use power rule } \\ 2x \ln(4)=(x-1) \ln(7) \\ 2 \ln(4) x=\ln(7) x- \ln(7) \]
now to get the x people together subtract ln(7)x on both sides
\[ax=bx-c \\ ax-bx=-c \\ x(a-b)=-c \\ \text{ the last step is to choose to divide by } \\ \text{ what is in front of the } x \\ \\ \text{ in this example that would be } (a-b) \\ x=\frac{-c}{a-b}\]
this is exactly what you are going to do here first subtract ln(7)x on both sides then use my example as a guide sorta to finally find x
2.77258872224x=1.94591014906x-1.94591014906
I thought you aren't allowed to approximate until you have final answer
i didn't :)
ln(7) is irrational there is no way you can write out the whole number same for ln(4)
oh. Sorry, I just typed it into my calculator & thats what it said
your calculator can only show you so many digits
it doesn't have an infinite screen display
anyways... \[\ln(4^{2x})=\ln(7^{x-1}) \\ \\ \text{ Now use power rule } \\ 2x \ln(4)=(x-1) \ln(7) \\ 2 \ln(4) x=\ln(7) x- \ln(7) \\ \text{ the step I was asking you \to do is subtract } \ln(7) x \text{ on both sides } \\ 2 \ln(4) x -\ln(7) x=-\ln(7)\] try factoring the x out on the left hand side like I did in my example ax-bx=x(a-b)
2 In(x)-In(-3)x=-In (7)
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