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

2 Answers
Apr 18, 2016

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

Apr 18, 2016

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.