Question

Question #070b2

Answer 1

Assuming that the logarithm is base 10:
`log_10(9) = ln(9)/ln(10)`

Answer 2

`log_10(9) = (ln(9))/(ln(10))`

Explanation:

Consider `log_10(9)`

Set `" " log_(10)(9) = x`

Another way of writing this is:`" "10^x=9`

Set `" "e^b=9=>color(red)(b=ln(9))`

But we have:`" " e^b=9=10^x`

Take natural logs of both sides and using the principle that
`ln(10^x)` is the same as `xln(10)`

`bln(e)=xln(10)`

but `ln(e)=1`

`b=xln(10)`

Thus `x=(color(red)(b))/(ln(10))`

but `b=ln(9)` giving

`x=ln(9)/(ln(10))`

But we set `log_(10)(9) = x` giving:

`log_10(9) = (ln(9))/(ln(10))`