Answer
Since $x \ne -5$, then $x = 2$
Work Step by Step
$\log x + \log (x+3) = \log 10$
$\log [x(x+3)] = \log 10$
$\log (x^{2} + 3x) = \log 10$
$x^{2} + 3x = 10$
$x^{2} + 3x - 10 = 0$
$(x+5)(x-2) = 0$
$x = -5, 2$
Since $x \ne -5$, then $x = 2$
College Algebra (6th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-32178-228-3
ISBN 13:
978-0-32178-228-1
Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.4 - Page 490: 87
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