How do you solve #log(y) = 1/4 log16 + 1/2 log49#?

Question

2550 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-02T07:41:21

#log y = 1/4 log 16 + 1/2 log 49 = log 2 + log 7 = log 14#

So #y = 14#

If #a > 0# then #log a^b = b log a#

If #a, b > 0# then #log ab = log a + log b#

So:

#log y = 1/4 log 16 + 1/2 log 49 = 1/4 log 2^4 + 1/2 log 7^2#

#= 1/4*4 log 2 + 1/2 * 2 log 7 = log 2 + log 7 = log (2*7)#

#= log 14#

That is: #log y = log 14#

If #log a = log b# then #a = b#

So #y = 14#