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# need help solving the equation?

log2 (x)+log2 (x-7)=3

### 3 Answers

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- intc_escapeeLv 71 decade agoFavorite Answer
log2(x)+log2(x-7)=3

x(x-7) = (2)³

(x - 8) (x + 1) = 0

x = 8, -1 .... discard -1

Answer: x = 8

- PurepotatoLv 41 decade ago
log2x + log2(x-7) = 3

log2 + logx + log2 + log(x-7) = 3

logx + log(x-7) = 3 - 2log2

log((x)(x-7)) = 3 - log(2^2)

10^(3 - log4) = (x)(x-7)

250 = x^2 - 7x

x^2 - 7x - 250 = 0

x = 19.69413474

x = -12.69413474 (Use Quadratic Formula)

x cannot be -12.69413474 because it doesn't work in the original equation; you cannot take the log of a negative number. Final answer: x=19.69413474

- ComoLv 71 decade ago
Let log be log to base 2

log [ ( x ) ( x - 7 ) ] = 3

[ ( x ) ( x - 7 ) ] = 2 ³

x² - 7x - 8 = 0

( x - 8 ) ( x + 1 ) = 0

x = 8 is acceptable

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