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

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Answer 1 · 2016-04-18T10:11:42

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

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 · 2016-04-18T13:46:41

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

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#.