How do you solve $Log x + log(x+3)= log 10$ ?

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Answer 1 · 2015-07-19T14:12:45

Use properties of $log$ , exponentiation and solve a quadratic to find solution $x=2$ .

$log(a)+log(b) = log(ab)$

So $log x + log(x+3) = log(x(x+3)) = log(x^2+3x)$

Note we require $x > 0$ in order that $log x$ is defined.

So:

$log(x^2+3x) = log 10$

Take exponent base $10$ of both sides to get:

$x^2 + 3x = 10$

Subtract $10$ from both sides to get:

$0 = x^2+3x-10 = (x+5)(x-2)$

So $x=-5$ or $x=2$ .

$x=-5$ is spurious, because we require $x > 0$ in order that $log x$ be defined.

So the only valid solution is $x=2$

How do you solve Log x + log(x+3)= log 10Log x + log(x+3)= log 10?