# How do you solve #1 + (2/ log_ (x-4)(x)) = log_ (x)((x-6)/6) + (1/ log_(x-6)(x))#?

##### 2 Answers

The given equation can be written as

is the problem correct?

The equation has no real solutions.

#### Explanation:

First, we should attempt to unify all the logarithms' bases: that is, make all the logarithms have the same base. To do this, rewrite the two terms that have a logarithm in the denominator using the change of base formula.

The change of base formula states that:

#log_a(b)=log_c(b)/log_c(a)#

Thus,

#log_(x-4)(x)=log_10(x)/log_10(x-4)=log(x)/log(x-4)#

The equation can then be rewritten as

#1+2/(log(x)/log(x-4))=log_x((x-6)/6)+1/(log(x)/log(x-6))#

The fractions in the denominator can be flipped (dividing a fraction is the same as multiplying by the fraction's reciprocal).

#1+(2log(x-4))/log(x)=log_x((x-6)/6)+log(x-6)/log(x)#

We can now use the change of base formula in reverse:

#log(b)/log(a)=log_a(b)#

Making the equation:

#1+2log_x(x-4)=log_x((x-6)/6)+log_x(x-6)#

On the left hand side, use the logarithm rule

#b*log(a)=log(a^b)#

Thus, we obtain the equation

#1+log_x[(x-4)^2]=log_x((x-6)/6)+log_x(x-6)#

Now, try to combine the logarithms.

On the right hand side, use the rule:

#log_c(a)+log_c(b)=log_c(ab)#

Yielding:

#1+log_x[(x-4)^2]=log_x((x-6)^2/6)#

Subtract the logarithm on the left from both sides.

#1=log_x((x-6)^2/6)-log_x[(x-4)^2]#

Now, use the rule:

#log_c(a)-log_c(b)=log_c(a/b)#

The logarithm is completely condensed:

#1=log_x((x-6)^2/(6(x-4)^2))#

Exponentiate both sides with base

#x^1=x^(log_x((x-6)^2/(6(x-4)^2))#

#x=(x-6)^2/(6(x-4)^2)#

#6x(x-4)^2=(x-6)^2#

#6x(x^2-8x+16)=x^2-12x+36#

#6x^3-48x^2+96x=x^2-12x+36#

#6x^3-49x^2+108x-36=0#

To solve this resulting cubic, we would have to use advanced solving techniques that are outside of the scope of this question, since all three roots are irrational. We can determine the roots approximately by examining a graph of

graph{6x^3-49x^2+108x-36 [-1, 6, -15, 43.5]}

The three roots are

**However,** none of these solutions are valid. If we plug any of these solutions into the original equation, we will have logarithms that do not function.

For example, in

Thus the equation has no real solutions.