**log**since we know that when the log

_{3}_{3}a = log

_{3}b, then very simply a must be equal to b. In this problem a=3x-6 and b=2x+1

**x=7**

solve the equation log3 (3x-6) = log3 (2x+1)

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George T. | George T.--"It's All About Math!"George T.--"It's All About Math!"

Heather:

In this problem, you shouldn't be confused by the **log**_{3}
since we know that when the log_{3} a = log_{3} b, then very simply a must be equal to b. In this problem a=3x-6 and b=2x+1

So: 3x-6=2x+1

3x-2x=1+6

1x=7

Hope this helps!

George T.

Hi Heather;

log_{3} (3x-6) = log_{3} (2x+1)

Before I answer this, I was like to briefly review logarithms.

Let's just say that...

log_{3} (3x-6) = 5

I randomly selected the number 5.

This would resolve as...

3x-6=3^{5}

Do you see how the base _{3} moved to the other side of the = sign and became a 3, whereas the 5 rose to exponential status of
^{5}?

I love the way Megan described the components as base, exponent and "answer". If she does not mind, I intend to use it in future answers.

In the equation you provided, the base of _{3} appears on both sides of the equation. Henceforth...

log_{3} (3x-6) = log_{3} (2x+1)

we can cancel these.

3x-6=2x+1

Let's add 6 to both sides as we proceed to isolate x...

3x-6+6=2x+1+6

3x=2x+7

Let's subtract 2x from both sides...

-2x+3x=2x+7-2x

x=7

Let's verify...

(3x-6) ??? (2x+1)

[3(7)-6] ??? [2(7)+1]

21-6 ??? 14+1

15=15

All Good!

Megan H. | Math Tutoring - All GradesMath Tutoring - All Grades

Hey Heather!

log_{3}9=2 -->example

3 is the base, 2 is the exponent and 9 is what I call the answer. You can rewrite this expression as 3^{2}=9 so you can see that. In the expression you have both of your logs have the same base of 3 and since they are set equal to each other tell you that they must have the same exponent. Therefore the answers must be equal so you can just set 3x-6=2x+1 and solve for x.

In this similar problem:

log2(2x-2)=log2(x+1)

2x-2 = x+1 -->set answers equal to eachother

x-2=1 --> solve for x

x=3

Good Luck!

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