Question

How do you rewrite `log_x(3/4)` as a ratio of common logs and natural logs?

Answer 1

In common logarithm `log_x (3/4)=log(3/4)/logx`

and in natural logarithm `log_x (3/4)=ln(3/4)/lnx`

Explanation:

Let `log_b a=x`, then we know that `b^x=a`

and taking common logs (i.e. to the base `10`) on both sides, we get

`logb^x=loga` i.e. `xlogb=loga` or `x=loga/logb`

As `x=log_b a`, we can say `log_b a=loga/logb`

Hence in common logarithm `log_x (3/4)=log(3/4)/logx`

Now as `b^x=a`, taking natural logarithm we get

`lnb^x=lna` i.e. `xlnb=lna` or `x=lna/lnb`

i.e. `log_x (3/4)=ln(3/4)/lnx`