How do you solve #2log_4x-log_4(x-1)=1#?

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Answer 1 · 2016-10-12T13:24:33

#x = 2#

Start by applying the rule #alogn = logn^a#.

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

Now apply the difference rules of logarithms: #log_a(n) - log_a(m) = log_a(n/m)#.

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

Convert to exponential form using the rule #log_a(n) = b -> a^b = n#

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

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

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

#x^2 = 4x - 4#

#x^2 - 4x + 4 = 0#

#(x - 2)(x - 2) = 0#

#x = 2#

Checking in the original equation:

#log_4(2^2) - log_4(2 - 1) =^? 1#

#log_4(4) - log_4(1) =^?1#

#1 - 0 = 1#

The solution works!

Hopefully this helps!