Exponential and Logarithmic Equations

[Koala]

Help! How do I solve this equation? 4^(1-2x)=2

This is what I've done:
Change to a logarithmic equation and move the exponent out in front: (1-2x)log(4)=log(2)
Distribute the (1-2x) through the log(4): log(4)+log(-8x)=log(2)
Subtract log(4) from both sides: log(-8x)=log(2)-log(4)
Use the log rule that allows us to separate out logs: log(-8)+log(x)=log(2)-log(4)
Subtract log(-8): log(x)=log(2)-log(4)-log(-8)
Change right side using properties of logarithms: log(x)=log(2/4)-log(-8)
Again: log(x)=log((2/4)/(-8))
Simplify: log(x)=log((1/2)/(-8))
Further simplify: log(x)=log(-1/16)

Is this right? What do I do to get rid of the log?

 

[lookagain]

Help! How do I solve this equation? 4^(1-2x)=2

Koala,

please do not try to do it that way. This problem
is geared toward the following:


4 and 2 are each a power of a common base, which is 2.


Write \(\displaystyle 4 \ as \ 2^2 \ \ and \ \ 2 \ \ as \ \ 2^1:\)


\(\displaystyle (2^2)^{(1 - 2x)} \ = \ 2^1\)


Multiply exponents together on the left-hand side:


\(\displaystyle 2^{(2 - 4x)} \ = \ 2^1\)


Because the sides of the equation are equal,
and the bases are equal, then equate the
exponents to each other:


2 - 4x = 1


Continue...

 

[Koala]

Thank you so much! That is so much easier!