SOLUTION: Please help me Solve for all values of x, to the nearest 10th. log4(2x−7)+log4(3x−3)=4

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Question 617078: Please help me
Solve for all values of x, to the nearest 10th.
log4(2x−7)+log4(3x−3)=4

Found 3 solutions by ewatrrr, jsmallt9, josmiceli:
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi

4^4 = (2x-7)(3x-3)
6x^2 -27x + 21 - 256 = 0
3x^2 - 27x -235 = 0

x: -4.7, 9.2 (rounded to the nearest tenth)

Answer by jsmallt9(3758)   (Show Source): You can put this solution on YOUR website!
Assuming these are base 4 logarithms...

We want the equation in the form
log(expression) = other-expression
So we will use a property of logarithms, , to combine the logarithms:


Now that the equation is in the desired form, the next step is to rewrite the equation in exponential form. In general is equivalent to . Using this pattern on our equation we get:

which simplifies to:


Now that the variable is out of the logarithm, we can solve for it. This is a quadratic equation so we want one side to be zero. Subtracting 256 from each side:

Next we factor or use the Quadratic Formula. This will not factor so we must use the formula:

Simplifying...





So or

Since you were asked for decimal answers, get out your calculator and round off your answer to the nearest tenth.

Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Use the general rule:

Also, express the right side as a log to base 4



















To nearest tenth,
check answer:






Looks close enough
-----------------
Note that had to be positive because a log
with a positive base (4), can't result in a negative number

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