How do you solve #2logx = log(-2x+15)#?

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Answer 1 · 2015-08-09T23:38:03

#color(red)(x=3)#

#2logx = log(-2x+15)#

Recall that #alog x= log(x^a)#, so

#logx^2 = log(-2x+15)#

Convert the logarithmic equation to an exponential equation.

#10^(logx^2) = 10^(log(-2x+15))#

Remember that #10^logx =x#, so

#x^2=-2x+15#

#x^2+2x-15 =0#

#(x+5)(x-3)=0#

#x+5=0# and #x-3=0#

#x=-5# and #x=3#

Check:

#2logx = log(-2x+15)#

If #x=-5#,

#2log(-5)= log(-2(-5)+15)#

This is impossible, because #log(-5)# is undefined.

If #x= 3#,

#2log3= log(-2×3+15)#

#log3^2= log(-6+15)#

#log9 = log9#

#x=3# is a solution.