How do you solve for x in $log x + log (3 x - 13) = 1$ $Log x + Log (3x-13)=1$ ?

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Answer 1 · 2018-07-20T04:23:32

$x = 5$ $x=5$ only

If the question just says $log$ $log$ , we can safely assume that it means ${log}_{10}$ $log_10$

${log}_{10} x + {log}_{10} (3 x - 13) = 1$ $log_10x+log_10(3x-13)=1$

${log}_{10} x (3 x - 13) = 1$ $log_10 x(3x-13)=1$
Recall: ${log}_{a} b + {log}_{a} c = {log}_{a} b c$ $log_a b+log_a c=log_a bc$

$x (3 x - 13) = 10^{1}$ $x(3x-13)=10^1$
Recall: If ${log}_{a} b = c$ $log_a b=c$ then $a^{c} = b$ $a^c=b$

$x (3 x - 13) = 10$ $x(3x-13)=10$

$3 x^{2} - 13 x - 10 = 0$ $3x^2-13x-10=0$

$(3 x + 2) (x - 5) = 0$ $(3x+2)(x-5)=0$

$x = - \frac{2}{3}$ $x=-2/3$ or $x = 5$ $x=5$

But $x = - \frac{2}{3}$ $x=-2/3$ is not a solution since you cannot log negative numbers so $x = 5$ $x=5$ is the solution only

How do you solve for x in logx+log(3x−13)=1Log x + Log (3x-13)=1?