Question

How do you solve `4^(3-2x)=5^-x`?

Answer 1

`x=((3log4)/(2log4-log5))≈3.5755`

Explanation:

`4^(3-2x)=5^-x`

1. Take the common logarithm of both sides
   `log4^(3-2x) = log5^-x`
2. Move the exponents in front of the logarithms
   `(3-2x)*log4 = -xlog5`
3. Distribute
   `3log4-2xlog4=-xlog5`
4. Rearrange the terms so that each "x" appears on the same side of the equation
   `3log4=2xlog4-xlog5`
5. Factor out x from the right side of the equation
   `3log4=x(2log4-log5)`
6. Isolate (solve for) the variable x by dividing both sides of the equation by `(2log4-log5)`.
   `(3log4)/(2log4-log5)=(xcancel((2log4-log5)))/(cancel(2log4-log5))`
7. (Optional) Plug into calculator to approximate a value for x
   `x=(3log4)/(2log4-log5)≈3.5755`