How do you solve for x in ${log}_{6} (2 - x) + {log}_{6} (3 - 4 x) = 1$ $log_ 6 (2-x) + log _ 6 (3-4x) = 1$ ?

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How do you solve for x in ${log}_{6} (2 - x) + {log}_{6} (3 - 4 x) = 1$ $log_ 6 (2-x) + log _ 6 (3-4x) = 1$ ?

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Answer 1 · 2016-02-06T05:47:47

$x = 0$ $x= 0$

Explanation:

Property of Logarithmic expression

$log A + log B = log (A B) (1)$ $log A + log B = Log(AB) " " " " " (1)$
$n log A = {log A}^{n} (2)$ $n log A= log A^n " " " " (2)$
${log}_{a} y = x \Rightarrow a^{x} = y (3)$ $log_ay= x => a^x = y " " " (3)$

Given :

${log}_{6} (2 - x) + {log}_{6} (3 - 4 x) = 1$ $log_6(2-x)+ log_6(3-4x) = 1$

Rewrite as:

Using rule (2)

${log}_{6} (2 - x) (3 - x) = 1$ $log_6(2-x)(3-x)= 1$

${log}_{6} (6 - 2 x - 3 x + x^{2}) = 1$ $log_6(6-2x-3x+x^2)= 1$

${log}_{6} (6 - 5 x + x^{2}) = 1$ $log_6(6-5x+x^2) = 1$

Using rule (3)

$6^{1} = 6 - 5 x^{2} + x^{2}$ $6^1 = 6-5x^2 + x^2$

Now we have a simple quadratic equation, let's solve it.

$x^{2} - 5 x + 6 = 6$ $x^2 -5x+6= 6$

$- 6 - 6$ $-6 " " " -6$

$= = = = = = = =$ $========$

$x^{2} - 5 x = 0$ $x^2 -5x = 0$
$x (x - 5) = 0$ $x(x-5) = 0$

$x = 0$ $x= 0$ or $x = 5$ $x= 5$

Always check your answer if you have 2 solutions when solving a log equation .

$C h e c k x = 5$ $color(blue)(Check " " x= 5)$

$x = 5$ $x= 5$

${log}_{6} (2 - 5) + {log}_{6} (3 - 4 (5)) = 1$ $log_6(2-5) + log_6(3-4(5))= 1$

${log}_{6} (- 3) + {log}_{6} (- 17) = 1$ $log_6(-3) + log_6(-17) = 1$

$x = 5 Extraneous s o l u t i o n$ $color(red)( x = 5 " Extraneous " solution)$

Can't evaluate the expression on the left, because the argument of any logarithm always POSITIVE and greater than zero, due to domain restriction

$C h e c k x = 0$ $color(blue)(Check " " x= 0)$

${log}_{6} (2 - 0) + {log}_{6} (3 - 4 (0)) = 1$ $color(red)(log_6(2-0)+log_6(3-4(0))= 1$

${log}_{6} (2) + {log}_{6} (3) = 1$ $log_6(2)+log_6(3) = 1$

${log}_{6} (6) = 1$ $log_6(6) = 1$

$1 = 1$ $1= 1$

Solution $x = 0$ $color(red)(x= 0)$

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How do you solve for x in ${log}_{6} (2 - x) + {log}_{6} (3 - 4 x) = 1$ $log_ 6 (2-x) + log _ 6 (3-4x) = 1$ ?

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Explanation:

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How do you solve for x in log6(2−x)+log6(3−4x)=1log_ 6 (2-x) + log _ 6 (3-4x) = 1?

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Explanation:

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How do you solve for x in ${log}_{6} (2 - x) + {log}_{6} (3 - 4 x) = 1$ $log_ 6 (2-x) + log _ 6 (3-4x) = 1$ ?