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

Mar 5, 2016

The given equation can be written as

$1 + 2 {\log}_{x} \left(x - 4\right) = {\log}_{x} \left(\frac{x - 6}{6}\right) + {\log}_{x} \left(x - 6\right)$
$\implies 1 = - 2 {\log}_{x} \left(x - 4\right) + {\log}_{x} \left(\frac{x - 6}{6}\right) + {\log}_{x} \left(x - 6\right)$
$\implies 1 = - {\log}_{x} {\left(x - 4\right)}^{2} + {\log}_{x} \left(\frac{x - 6}{6}\right) + {\log}_{x} \left(x - 6\right)$
$\implies 1 = {\log}_{x} \left({\left(x - 6\right)}^{2} / \left(6 {\left(x - 4\right)}^{2}\right)\right)$
$\implies x = \left({\left(x - 6\right)}^{2} / \left(6 {\left(x - 4\right)}^{2}\right)\right)$
$\implies 6 x {\left(x - 4\right)}^{2} = {\left(x - 6\right)}^{2}$
$6 {x}^{3} - 48 {x}^{2} + 96 x - {x}^{2} + 12 x - 36 = 0$
$6 {x}^{3} - 49 {x}^{2} + 108 x - 36 = 0$

is the problem correct?

Mar 5, 2016

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:

${\log}_{a} \left(b\right) = {\log}_{c} \frac{b}{\log} _ c \left(a\right)$

Thus,

${\log}_{x - 4} \left(x\right) = {\log}_{10} \frac{x}{\log} _ 10 \left(x - 4\right) = \log \frac{x}{\log} \left(x - 4\right)$

The equation can then be rewritten as

$1 + \frac{2}{\log \frac{x}{\log} \left(x - 4\right)} = {\log}_{x} \left(\frac{x - 6}{6}\right) + \frac{1}{\log \frac{x}{\log} \left(x - 6\right)}$

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

$1 + \frac{2 \log \left(x - 4\right)}{\log} \left(x\right) = {\log}_{x} \left(\frac{x - 6}{6}\right) + \log \frac{x - 6}{\log} \left(x\right)$

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

$\log \frac{b}{\log} \left(a\right) = {\log}_{a} \left(b\right)$

Making the equation:

$1 + 2 {\log}_{x} \left(x - 4\right) = {\log}_{x} \left(\frac{x - 6}{6}\right) + {\log}_{x} \left(x - 6\right)$

On the left hand side, use the logarithm rule

$b \cdot \log \left(a\right) = \log \left({a}^{b}\right)$

Thus, we obtain the equation

$1 + {\log}_{x} \left[{\left(x - 4\right)}^{2}\right] = {\log}_{x} \left(\frac{x - 6}{6}\right) + {\log}_{x} \left(x - 6\right)$

Now, try to combine the logarithms.

On the right hand side, use the rule:

${\log}_{c} \left(a\right) + {\log}_{c} \left(b\right) = {\log}_{c} \left(a b\right)$

Yielding:

$1 + {\log}_{x} \left[{\left(x - 4\right)}^{2}\right] = {\log}_{x} \left({\left(x - 6\right)}^{2} / 6\right)$

Subtract the logarithm on the left from both sides.

$1 = {\log}_{x} \left({\left(x - 6\right)}^{2} / 6\right) - {\log}_{x} \left[{\left(x - 4\right)}^{2}\right]$

Now, use the rule:

${\log}_{c} \left(a\right) - {\log}_{c} \left(b\right) = {\log}_{c} \left(\frac{a}{b}\right)$

The logarithm is completely condensed:

$1 = {\log}_{x} \left({\left(x - 6\right)}^{2} / \left(6 {\left(x - 4\right)}^{2}\right)\right)$

Exponentiate both sides with base $x$ to undo the logarithm, then solve.

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

$x = {\left(x - 6\right)}^{2} / \left(6 {\left(x - 4\right)}^{2}\right)$

$6 x {\left(x - 4\right)}^{2} = {\left(x - 6\right)}^{2}$

$6 x \left({x}^{2} - 8 x + 16\right) = {x}^{2} - 12 x + 36$

$6 {x}^{3} - 48 {x}^{2} + 96 x = {x}^{2} - 12 x + 36$

$6 {x}^{3} - 49 {x}^{2} + 108 x - 36 = 0$

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 $6 {x}^{3} - 49 {x}^{2} + 108 x - 36$.

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

The three roots are $x \approx 0.404 , 3.435 , 4.382$.

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 ${\log}_{x - 6} \left(x\right)$, all of the $x$-values will cause the value of the base to be negative. For ${\log}_{a} \left(b\right)$ to be defined, $a , b \ge 0$.

Thus the equation has no real solutions.