SOLUTION: log base 2 (x+1) = log base 8 (3x). solve x.

Question 270091: log base 2 (x+1) = log base 8 (3x). solve x.
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!

If both of the logarithms had the same base we could simply say
x+1 = 3x
and solve it. But the bases are not the same, yet. We will start by changing the base 8 logarithm to an expression of base 2 logarithms. The formula for changing bases is: . Using this to change the base 8 log to an expression of base 2 logs we get:

And since then . Substituting this into our equation we get:

Next we need to get rid of the fraction so we'll multiply both sides by 3:

Next the 3 needs to be moved. We have a property of logarithms, , which allows us to move a coefficient into the argument as an exponent. Using this on your equation we get:

We now have an equation that says that two base 2 logarithms are equal. And if the logs are equal then their arguments must be equal:

This is an equation we could try to solve. Cubing the left side we get:

Subtracting 3x from each side we get:

Normally we would try to factor the left side. But there is no method of factoring I know of that will factor it. So we are stuck. (This makes we wonder if you posted the question correctly.)

If we graph and look for where the graph crosses the x-axis (where y is zero), we can make a guess at the solution:

From this graph we can see that our solution is a number somewhere near -3.1 or -3.2
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