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You calculate a log, you do not solve a log!
log(9x) + log(x) = 4log(10)log(9) + log(x) + log(x) = 4log(10)2log(x) = 4log(10) - log(9)log(x2) = log(104) - log(9)log(x2) = log(104/9)x2 = 104/9x = 102/3x = 33 and 1/3
Sometimes you need to take logs, or antilogs, on both sides of an equation. Sometimes you need to apply certain common logarithmic identities, especially: log(xy) = log x + log y log (x/y) = log x - log y log (ab) = b log a