How do you sole X when logx-3 plus logx-2 equals log2x plus 24?
We can't see the parentheses, and there are at least two ways to read this.Here are solutions for the most likely two:--------------------------------------------------------log(x) - 3 + log(x) - 2 = log(2x) + 24Add 5 to each side:log(x) + log(x) = log(2x) + 29Subtract log(2x) from each side:log(x) + log(x) - log(2x) = 29Combine the logs on the left side, and massage:log( x2/2x ) = log( x/2 ) = 29Take the antilog of each side:x/2 = 1029Multiply each side by 2:x = 2 x 1029------------------------------------------------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 + 24Expand the left side:x2 - 5x +6 = 2x + 24Subtract (2x+24) from each side:x2 - 7x - 18 = 0Factor:(x - 9) (x + 2) = 0Whence:x = 9x = -2We have to discard the solution [ x = -2 ] because one term in the equationis log(x-2).If 'x' were -2 then we'd have log(-4) but negative numbers don't have logs.