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#