How do you solve ${log}_{27} x = 1 - {log}_{27} (x - 0.4)$ $Log_27x = 1 - log_27 (x - 0.4)$ ?

Question

6034 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-09T05:06:11

$x = 5.4$ $x = 5.4$

Use ${log}_{b} m + {log}_{b} n = {log}_{b} (m n)$ $log_b m+log_b n= log_b(mn)$ , and

if $c = {log}_{b} a, a = b^{c}$ $c=log_b a, a=b^c$

Here, ${log}_{27} x + {log}_{27} (x - 0.4) = {log}_{27} (x (x - 0.4)) = 1$ $log_27 x+log_27(x-0.4)=log_27(x(x-0.4))=1$ .

So, $x (x - 0.4) = 27^{1} = 27$ $x(x-0.4)=27^1=27$

$x^{2} - 0.4 x - 27 = 0$ $x^2-0.4x-27=0$ . Solving.

$x = 5.4 and - 5$ $x = 5.4 and -5$

x > 0.4. So, x = 5.4.

How do you solve log27x=1−log27(x−0.4)Log_27x = 1 - log_27 (x - 0.4)?