Question

Solve for x. e^4x=(1/64) so first I can do ln e ^4x=ln(1/64) right? I am then having trouble from here. Please help.

Answer 1

you need to take the base e log of both sides

so you have

\[\ln(e^{4x}) = \ln(\frac{1}{64})\]

Answer 2

the left hand side of the equation becomes 4x

so you are solving

\[4x = \ln(\frac{1}{64})\]

you need to use the lag law for division

and the law for ln(1)

to get the answer

Answer 3

When you say log law for division are you talking about log base b(x)-log base b (y)?

Answer 4

on the right hand side you are dividing 1 by 64

so you may choose to use

4x = ln(1) - ln(64)

and then work from there

Answer 5

but you need to know ln(1)

Answer 6

Ok so ln base e (1)=1 right? So I did 4x-1=ln base e 64

Answer 7

ok wait so ln(1) is not the same as ln base e(1) correct?

Answer 8

ln(1)=0 ?

Answer 9

not quite ln(1) = 0

so you would have

4x = 0 - ln(64)

solve for x

Answer 10

Use the property \[\log_{a}(a)^p = p \] ,,,,
that is , when base and the final number are same, take power as the value..

Answer 11

But I don't see the base and final number being the same. Right now I am at x=-ln(64)/4

Answer 12

this is what you need to solve

\[4x = - \ln(64)\]

just solve for x

Answer 13

I meant that \[\log_{e} (e)^(4x) = 4x\]
And then, you can directly write
\[\log_{e} (1/64) = \ln (4)^(-3)\]