How do you solve $log (3 x + 7) + log (x - 2) = 1$ $log(3x + 7)+ log(x - 2)= 1$ ?

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How do you solve $log (3 x + 7) + log (x - 2) = 1$ $log(3x + 7)+ log(x - 2)= 1$ ?

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Answer 1 · 2016-03-16T22:10:21

$x = \frac{8}{3}$ $x=8/3$

Explanation:

$1$ $1$ . Use the log property, ${log}_{b} (m \cdot n) = {log}_{b} (m) + {log}_{b} (n)$ $log_color(purple)b(color(red)m*color(blue)n)=log_color(purple)b(color(red)m)+log_color(purple)b(color(blue)n)$ to simplify the left side of the equation.

$log (3 x + 7) + log (x - 2) = 1$ $log(3x+7)+log(x-2)=1$

$log ((3 x + 7) (x - 2)) = 1$ $log((3x+7)(x-2))=1$

$log (3 x^{2} + x - 14) = 1$ $log(3x^2+x-14)=1$

$2$ $2$ . Use the log property, ${log}_{b} (b^{x}) = x$ $log_color(purple)b(color(purple)b^color(orange)x)=color(orange)x$ , to rewrite the right side of the equation.

$log (3 x^{2} + x - 14) = log (10)$ $log(3x^2+x-14)=log(10)$

$3$ $3$ . Since the equation now follows a " $log = log$ $log=log$ " situation, where the bases are the same on both sides, rewrite the equation without the "log" portion.

$3 x^{2} + x - 14 = 10$ $3x^2+x-14=10$

$4$ $4$ . Subtract $10$ $10$ from both sides of the equation.

$3 x^{2} + x - 24 = 0$ $3x^2+x-24=0$

$5$ $5$ . Factor the quadratic equation.

$(3 x - 8) (x + 3) = 0$ $(3x-8)(x+3)=0$

$6$ $6$ . Set each factor to $0$ $0$ and solve for $x$ $x$ .

$3 x - 8 = 0 X X X X X X X x + 3 = 0$ $3x-8=0color(white)(XXXXXXX)x+3=0$

$3x=8color(white)(XXXXXXXXX)color(red)cancelcolor(green)(|bar(ul(color(white)(a/a)x=-3color(white)(a/a)|)))$

$color(green)(|bar(ul(color(white)(a/a)x=8/3color(white)(a/a)|)))$

$7$ . However, $color(red)(x=-3)$ does not satisfy the equation since it produces a log with a negative number, and you $color(teal)"can't log a negative number"$ .

For example, when you substitute $color(red)(x=-3)$ back into the original equation:

$log(3x+7)+log(x-2)=1$

$log(3color(red)((-3))+7)+log(color(red)((-3))-2)=1$

$log(-9+7)+log(-5)=1$

$color(teal)(log(-2))+color(teal)(log(-5))=1$

For this reason, $color(red)(x=-3)$ is not a solution of the equation.

$:.$ , $x=8/3$ .

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How do you solve $log (3 x + 7) + log (x - 2) = 1$ $log(3x + 7)+ log(x - 2)= 1$ ?

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How do you solve log(3x + 7)+ log(x - 2)= 1log(3x+7)+log(x−2)=1log(3x + 7)+ log(x - 2)= 1?

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How do you solve $log (3 x + 7) + log (x - 2) = 1$ $log(3x + 7)+ log(x - 2)= 1$ ?