Question 1202488
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Answers: {{{x = 5/2 }}} and {{{ x = 2/3}}}


Work Shown:


All logs for this problem are base 7
log(3x) + log(2x-1) = log(16x-10)
log(3x*(2x-1)) = log(16x-10) ..... log rule log(A)+log(B) = log(A*B)
3x*(2x-1) = 16x-10
6x^2-3x = 16x-10
6x^2-3x-16x+10 = 0
6x^2-19x+10 = 0


Now use the quadratic formula
a = 6
b = -19
c = 10
{{{x = (-b +- sqrt(b^2 - 4ac))/(2a)}}}


{{{x = (-(-19) +- sqrt((-19)^2 - 4(6)(10)))/(2(6))}}}


{{{x = (19 +- sqrt(361-240))/(2(6))}}}


{{{x = (19 +- sqrt(121))/(12)}}}


{{{x = (19 +- 11)/(12)}}}


{{{x = (19 + 11)/(12) }}} or {{{ x = (19 - 11)/(12)}}}


{{{x = (30)/(12) }}} or {{{ x = (8)/(12)}}}


{{{x = 5/2 }}} or {{{ x = 2/3}}}


From here we need to check each possible solution.
I'll let the student do this part. Plug each x value one at a time into the original equation. Simplify both sides.
You should find that both x values work.
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