Question

(Exponential Equation) How do I find X?

`5^x=4^(x+1)`

Answer 1

`x~~ 6.21`.

Explanation:

Take the natural logarithm of both sides.

`ln(5^x) = ln(4^(x +1))`

Now use `lna^n = nlna`.

`xln5 = (x + 1)ln4`

`xln5 = xln4 + ln4`

`xln5 - xln4 = ln4`

`x(ln5 - ln4) = ln4`

This can be simplified further using `lna -lnb = ln(a/b)`.

`x(ln(5/4)) = ln4`

`x = (ln4)/(ln(5/4))`

If you prefer an approximation, we can take `x~~ 6.21`.

Hopefully this helps!

Answer 2

I got: `x=(ln(4))/(ln(5)-ln(4))`

Explanation:

Here we can try applying the natural log, `ln`, on both sides and apply some properties of logs:
`ln(5)^x=ln(4)^(x+1)`
then:
`xln(5)=(x+1)ln(4)`
rearrange:
`xln(5)=xln(4)+ln(4)`
`xln(5)-xln(4)=ln(4)`
`x[ln(5)-ln(4)]=ln(4)`
`x=(ln(4))/(ln(5)-ln(4))`
if you have a pocket calculator we can easily evaluate the natural log to get:
`x=6.21256`