Question

How do you solve `2^x*5=10^x`?

Answer 1

If `A = B`, then `logA = logB`:

`log(2^x xx 5) = log(10^x)`

Use the rule `log_a(n/m) = log_an - log_am` to simplify.

`log2^x + log5 = log10^x`

Use the rule `loga^n = nloga` to simplify further.

`xlog2 + log5 = xlog10`

`log5 = xlog10 - xlog2`

Factor out the `x`.

`log5 = x(log10 - log2)`

Put back the logs on the left side into quotient form to simplify.

`log5 = x(log(10/2))`

`log5 = x(log5)`

Isolate x.

`log5/log5 = x`

`1 = x`

Checking in the original equation, we find that this solution works.

Hopefully this helps!

Answer 2

`x=1` (using alternate solution method)

Explanation:

If `2^x*5=10^x`

`rArr 2^x*5 =(2*5)^x`

`rArr 2^x*5 = 2^x*5^x`

`rArr 5=5^x`

`rArr x=1`