1. Use the log property, 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(3x+7)+log(x-2)=1
log((3x+7)(x-2))=1
log(3x^2+x-14)=1
2. Use the log property, log_color(purple)b(color(purple)b^color(orange)x)=color(orange)x, to rewrite the right side of the equation.
log(3x^2+x-14)=log(10)
3. Since the equation now follows a "log=log" situation, where the bases are the same on both sides, rewrite the equation without the "log" portion.
3x^2+x-14=10
4. Subtract 10 from both sides of the equation.
3x^2+x-24=0
5. Factor the quadratic equation.
(3x-8)(x+3)=0
6. Set each factor to 0 and solve for x.
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.
