How do you solve #4^(log_4(x+8))=4^2#?

Question

5649 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-06T18:23:06

Soln. is #x=8.#

Given that, #4^(log_4^(x+8))#=4^2#

Since the bases are same, the indices must be equal, and hence,

#log_4^(x+8)=2#
#rArr x+8=4^2=16#

# rArr x=8#

Verification holds good, so the Soln. is #x=8.#

Answer 2 · 2016-07-06T18:27:26

see below

lots of ways to do this

#4^(log_4(x+8))=4^2#

you can see immediately that #x+8 = 4^2 implies x = 8#

this is because: #a ^ {ln_a Q} = Q#

or you can plod through mechanically. So, just equating the exponents....

#log_color{red}{4}(x+8)=2 qquad star#

#(x+8)=color{red}{4}^2# same result

or changing the base on the LHS of #star# to 2

#log_color{red}{4}(x+8) = (log_2(x+8))/(log_2 4)#

so

#(log_2(x+8))/(log_2 4)=2#

#implies log_2(x+8)=2* log_2 4#

#implies log_2(x+8)=2* 2 = 4#

#implies (x+8)=2^4 = 16 implies x = 8#