How do you solve #\log _ { 4} x -1/ 2\log _ { 4} ( x - 4) = 1#?

2 Answers
Noah G
Feb 25, 2017

#{8}#

Explanation:

You're going to have to get rid of the coefficient #1/2# by using #alogn = logn^a#.

#log_4 x - log_4 (x - 4)^(1/2) = 1#

#log_4 x - log_4 sqrt(x - 4) = 1#

Now use the fact that #log_a n - log_a m = log_a (n/m)#.

#log_4 (x/sqrt(x - 4)) = 1#

Convert to exponential form using #log_a n = m -> a^m = n#

#x/sqrt(x - 4) = 4^1#

#x = 4sqrt(x - 4)#

#x^2= (4sqrt(x - 4))^2#

#x^2 = 16(x- 4)#

#x^2 = 16x - 64#

#x^2 - 16x + 64 = 0#

#(x - 8)(x- 8) = 0#

#x = 8#

Check your result to ensure that the solution isn't extraneous.

#log_4 8 - 1/2log_4(8 - 4) =^? 1#

#3/2 - 1/2(1) = 1 color(green)(√)#

Hopefully this helps!

P dilip_k
Feb 25, 2017

#x=8#

Explanation:

#\log _ { 4} x -1/ 2\log _ { 4} ( x - 4) = 1#

#=>log _ { 4} x -log _ { 4} ( x - 4)^(1/2) =log_4 4#

#=>log _ { 4} (x/( x - 4)^(1/2)) =log_4 4#

#=> (x/( x - 4)^(1/2)) =4#

#=> (x^2/( x - 4)) =4^2=16#

#=>x^2=16x-64#

#=>x^2-16x+64=0#

#=>x^2-2*x*8+8^2=0#

#=>(x-8)^2=0#

#=>x=8#

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