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.

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