SOLUTION: Log7(3x)+log7(2x-1)=log7(16x-10)

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Question 1202488: Log7(3x)+log7(2x-1)=log7(16x-10)
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Answers: x+=+5%2F2+ and +x+=+2%2F3

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+=+%28-b+%2B-+sqrt%28b%5E2+-+4ac%29%29%2F%282a%29

x+=+%28-%28-19%29+%2B-+sqrt%28%28-19%29%5E2+-+4%286%29%2810%29%29%29%2F%282%286%29%29

x+=+%2819+%2B-+sqrt%28361-240%29%29%2F%282%286%29%29

x+=+%2819+%2B-+sqrt%28121%29%29%2F%2812%29

x+=+%2819+%2B-+11%29%2F%2812%29

x+=+%2819+%2B+11%29%2F%2812%29+ or +x+=+%2819+-+11%29%2F%2812%29

x+=+%2830%29%2F%2812%29+ or +x+=+%288%29%2F%2812%29

x+=+5%2F2+ or +x+=+2%2F3

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.