How do you sole X when logx-3 plus logx-2 equals log2x plus 24?

Answer 1

We can't see the parentheses, and there are at least two ways to read this.

Here are solutions for the most likely two:

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log(x) - 3 + log(x) - 2 = log(2x) + 24

Add 5 to each side:

log(x) + log(x) = log(2x) + 29

Subtract log(2x) from each side:

log(x) + log(x) - log(2x) = 29

Combine the logs on the left side, and massage:

log( x2/2x ) = log( x/2 ) = 29

Take the antilog of each side:

x/2 = 1029

Multiply each side by 2:

x = 2 x 1029

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log(x - 3) + log(x - 2) = log(2x + 24)

Combine logs on the left side:

log[ (x-3) (x-2) ] = log(2x + 24)

Take antilog of each side:

(x-3) (x-2) = 2x + 24

Expand the left side:

x2 - 5x +6 = 2x + 24

Subtract (2x+24) from each side:

x2 - 7x - 18 = 0

Factor:

(x - 9) (x + 2) = 0

Whence:

x = 9

x = -2

We have to discard the solution [ x = -2 ] because one term in the equation

is log(x-2).

If 'x' were -2 then we'd have log(-4) but negative numbers don't have logs.