`log(2x+1)-log(3x-2)=1`

To solve, first express the left side as one logarithm. To do so, apply the quotient rule which is `log_bM/N=log_bM-log_bN` .

`log((2x+1)/(3x-2))=1`

Then, express the equation to its equivalent exponential form to eliminate the logarithm.

Note that the exponential form of `log_b M=a` is `M=b^a` .

Since the base...

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`log(2x+1)-log(3x-2)=1`

To solve, first express the left side as one logarithm. To do so, apply the quotient rule which is `log_bM/N=log_bM-log_bN` .

`log((2x+1)/(3x-2))=1`

Then, express the equation to its equivalent exponential form to eliminate the logarithm.

Note that the exponential form of `log_b M=a` is `M=b^a` .

Since the base of the logarithm in the given equation is not written, it indicates that its base is 10.

So re-writing the equation, it becomes:

`log_10((2x+1)/(3x-2))=1`

And its exponential form is:

`(2x+1)/(3x-2)=10^1`

`(2x+1)/(3x-2)=10`

Now that the equation has no more logarithm, the next step is to remove the x in the denominator.

To do so, multiply both sides by 3x-2.

`(3x-2)*(2x+1)/(3x-2)=10*(3x-2)`

`2x+1=30x-20`

Next, combine like terms.

To combine 30x and 2x, bring them together on one side of the equation. So, move 2x to the right side by subtracting both sides by 2x.

`2x-2x+1=30x-2x-20`

`1=28x-20`

To combine 20 and 1, bring them together on the side opposite the term with x. So, add both sides by 20.

`1+20=28x-20+20`

`21=28x`

And, divide both sides by 28 to have x only at the right side.

`21/28=(28x)/28`

`3/4=x`

**Hence, the solution to the given equation is `x=3/4` .**