How do you solve #Log(x+2)+Log(x-1)=Log(88)#?

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Answer 1 · 2015-12-15T02:39:10

#x=9#

Use the product rule of logarithms: #log(a)+log(b)=log(ab)#

Thus, the expression can be written as

#log((x+2)(x-1)=log(88)#

Distribute.

#log(x^2+x-2)=log(88)#

Raise both sides as the power of #10#.

#10^(log(x^2+x-2))=10^(log(88))#

#x^2+x-2=88#

#x^2+x-90=0#

#(x+10)(x-9)=0#

#x+10=0#
or
#x-9=0#

#x=-10#
or
#x=9#

Plug in both potential values.

Notice that if you plug in #-10#, you'd have to take the logarithm of a negative number, which is impossible.

Thus, the #-10# answer is thrown out and all that's left is

#x=9#