Question #072fe

2 Answers
Jan 23, 2017

Answer:

Please see the explanation.

Explanation:

I am going to assume base 10:

#log_10(root(3)(10(x-1)^2)) - 1/3log_10((x+3)^2) = 1/3#

Remove the root from inside the first logarithm by multiplying by 1/3:

#1/3log_10(10(x-1)^2) - 1/3log_10((x+3)^2) = 1/3#

Multiply both sides of the equation by 3:

#log_10(10(x-1)^2) - log_10((x+3)^2) = 1#

The difference of two logarithms is the same as division within the argument:

#log_10(10(x-1)^2/(x+3)^2) = 1#

Eliminate the logarithm by making both sides the power of 10:

#10^(log_10(10(x-1)^2/(x+3)^2)) = 10^1#

On the left side the base and the logarithm are removed:

#10(x-1)^2/(x+3)^2 = 10^1#

Divide both sides by 10:

#(x-1)^2/(x+3)^2 = 1#

Multiply both sides by #(x+3)^2#:

#(x-1)^2=(x+3)^2#

Expand the squares:

#x^2 - 2x + 1 = x^2 + 6x + 9#

Add #-x^2 -6x - 1# to both sides:

#-8x = 8#

Divide by sides by -8:

#x = -1#

Jan 23, 2017

Answer:

#x=-5/4#

Explanation:

#logroot3(10(x-1)^2)-1/3log(3+x)^2=1/3#

#logroot3(10(x-1)^2)-log(3+x)^(2/3)#

#log(root3(10(x-1)^2)/(3+x)^(2/3))=1/3#

#root3(10(x-1)^2)/(3+x)^(2/3)=10^(1/3)#

#root3(10(x-1)^2)=10^(1/3)((3+x)^(2/3))#

#(root3(10(x-1)^2))^3=(10^(1/3)((3+x)^(2/3)))^3#

#10(x-1)^2=10(3+x)^2#

#(x-1)^2=(3+x)^2#

#x^2-2x-1=x^2+6x+9#

#8x+10=0#

#8x=-10#

#x=-10/8=-5/4#