Question

How do you solve `10^x=5^(x+1)`?

Answer 1

`x=log 5/log 2`.

Explanation:

`10^x=(2 X 5)^x=2^x5^x=5^(x+1)=5^x5^1`.

Cancelling the common factor `5^x`,

`2^x=5`.

Equating the logarithms,

`x log 2 = log 5`.

So, `x=log 5/log 2`.

Answer 2

I found: `x=2.32193`

Explanation:

We can try taking the natural log of both sides and apply one property of logs to write:
`ln(10^x)=ln(5^(x+1))`
and then for the exponents:
`xln(10)=(x+1)ln(5)`
rearrange:
`xln(10)=xln(5)+ln(5)`
`xln(10)-xln(5)=ln(5)`
so:
`x[ln(10)-ln(5)]=ln(5)`
and:
`x=(ln(5))/(ln(10)-ln(5))=ln5/ln(10/5)=ln5/ln2=2.32193`