Given log(base 4) m=a and log(base 12) m = b,

Find log(base 48) 3 in terms of a and b. Please help, thanks!

Answer: (a-b)/(a+b)

Use the change of base formula and log properties:
Log(b)(a)= loga/ logb in any base
Log(ab)= loga + logb
Log(a/b)= loga -logb
----------

Given log(base 4) m=a and log(base 12) m = b
Find log(base 48) 3 in terms of a and b.

Log(48)[3] = Log(m)[3] / log(m)[ 48]=

log(m)[12/4] / log(m)[12*4]=

log(m)[12] - log(m)[4]
-------------------------------=
Log(m)[12 ] + log(m)[4]

b-a
------
a+ b


I hope this helps!