Question 771261
<pre>
The other tutor, Alan, is right about your equation.

The bases 2 and 3 are not powers of the same number. 

However equations of the form 

log<sub>A</sub>(Bx+C) = log<sub>D</sub>(Ex+F)

can be solved algebraically if the bases are 
powers of the same integer, such as this equation,

log<sub>4</sub>(2x-1) = log<sub>8</sub>(7x-8)

This is solvable by algebraic methods:

Let both sides equal to K

log<sub>4</sub>(2x-1) = K  and  log<sub>8</sub>(7x-8) = K

4<sup>K</sup> = 2x-1       and  8<sup>K</sup> = 7x-8

(2<sup>2</sup>)<sup>K</sup> = 2x-1     and  (2<sup>3</sup>)<sup>K</sup> = 7x-8

2<sup>2K</sup> = 2x-1       and  2<sup>3K</sup> = 7x-8

Raise both sides of the first equation to the 3rd power and both sides
of the second equation to the 2nd power.

(2<sup>2K</sup>)<sup>3</sup> = (2x-1)<sup>3</sup>     and  (2<sup>3K</sup>)<sup>2</sup> = (7x-8)<sup>2</sup>

2<sup>6K</sup> = (2x-1)<sup>3</sup>     and  2<sup>6K</sup> = (7x-8)<sup>2</sup>

Now we can equate the rights sides since both equal2<sup>6K</sup>

(2x-1)<sup>3</sup> = (7x-8)<sup>2</sup>

   8x³ - 12x² + 6x - 1 = 49x²- 112x + 64

8x³ - 61x² + 118x - 65 = 0

Find possible rational zeros. 

±1, {{{"" +- 1/2}}}, {{{"" +- 1/4}}}, {{{"" +- 1/8}}}, ±5, {{{"" +- 5/2}}}, {{{"" +- 5/4}}}, {{{"" +- 5/8}}}, ±13, {{{"" +- 13/2}}}, {{{"" +- 13/4}}}, {{{"" +- 13/8}}}, ±65  

Try 1 with synthetic division:

1 | 8  -61   118  -65
  |<u>      8   -53   65</u>
    8  -53    65    0

So we have factored the equation as

(x-1)(8x²-53x+65) = 0

And the quadratic factors as

(x-1)(x-5)(8x-13) = 0

x = 1, x = 5, and x = {{{13/8}}}

So there are three answers, but we must check
to see if any are extraneous.  1 does not
check because it causes 7x-8 to be negative.
The other two solutions check.
So there are two solutions: 

x = 5, and x = {{{13/8}}}

But that was solvable only because the bases were both
powers of 2.
  | 
Edwin</pre>