How do you solve #Logx + Log(x+9)=1#?

2 Answers
May 16, 2018

See below

Explanation:

Our goal is to have an expresion like this #logA=logB# due to one to one aplication of logarithm, we get #A=B#. We have to apply the rules of logarithm for product, quotient and power and use the fact #log10=1#

#logx+log(x+9)=1#
#logx(x+9)=log 10#

Now, we know that #x(x+9)=10#
#x^2+9x-10=0# using quadratic formula we get

#x=-10 and x=1# reject negative because there is no logarithm and the only solution is #x=1#. Lets check it

#log1+log(1+9)=0+1=1#

May 16, 2018

#x=1#

Explanation:

#logx+log(x+9)=1#

#=>log(x(x+9))=1#

i.e. #x(x+9)=10#

or #x^2+9x-10=0#

or #x^2-x+10x-10=0#

or #x(x-1)+10(x-1)=0#

or #(x-1)(x+10)=0#

so either #x-1=0# i.e. #x=1#

or #x+10=0# i.e. #x=-10#

but we cannot have #x<=-9# as #x+9>0#

Hence only solution is #x=1#