How do you solve #Log_27x = 1 - log_27 (x - 0.4)#?

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Answer 1 · 2016-05-09T05:06:11

#x = 5.4#

Use #log_b m+log_b n= log_b(mn)#, and

if #c=log_b a, a=b^c#

Here, #log_27 x+log_27(x-0.4)=log_27(x(x-0.4))=1#.

So, #x(x-0.4)=27^1=27#

#x^2-0.4x-27=0#. Solving.

#x = 5.4 and -5#

x > 0.4. So, x = 5.4.