Question

How do you write `log_6 5` as a logarithm of base 4?

Answer 1

`log_(6)5=0.7737xxlog_(4)5`

Explanation:

Let `log_xa=p` and `log_cx=q`.

i.e. `x^p=a` and `c^q=x` and hence `a=(c^q)^p=c^(pq)`

i.e. `log_ca=pxxq` or `log_ca=log_xaxxlog_cx`-------(A)

Hence `log_(6)5=log_(4)5xxlog_(6)4`...........(B)

(A) also tells us that `log_xa=log_ca/log_cx`

and hence `log_(6)4=log_(10)4/log_(10)6` and putting this in (B)

`log_(6)5=log_(4)5xxlog4/log6=0.6021/0.7782xxlog_(4)5=0.7737xxlog_(4)5`

Answer 2

Use the change of base formula or solve an equation using log base 4.

Explanation:

Change of base formula

`log_b x = log_cx/log_cb`

So `log_6 5 = log_4 5/log_4 6`

Solve an equation

If you don't remember the change of base formula (or if you want to see where it comes from)

Let `x = log_6 5`

So `6^x=5`

Because we want log base 4, take that log on both sides:

`log_4(6^x)=log_4 5`

Now use the exponent property of logarithms:

`xlog_4 6=log_4 5`.

Finally divide by `log_4 6` to get

`x=log_4 5/log_4 6`.