How do you solve #log(x+3)+log(x)=1#?

1 Answer
Bdub
Mar 2, 2016

#log_10(x+3)+log_10(x)=1->log_10((x+3)(x))=1->log_10x^2+3x=1->10^1=x^2+3x#
#0=x^2+3x-10->0=(x+5)(x-2)->x=-5,x=2#

Explanation:

Use the property of logarithm for log of a product is the sum of the logs to change the sum to a product. Then use the definition to change from logarithmic form to exponential form then solve for x.

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