# Solve: `log_3 6 + log_3 (3x+1) = 2`

justaguide | College Teacher | (Level 2) Distinguished Educator

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The equation `log_3 6 + log_3 (3x+1) = 2` has to be solved for x.

Use the property of logarithm log a + log b = log (a*b)

`log_3 6 + log_3 (3x+1) = 2`

=> `log_3 (6*(3x + 1)) = 2`

=> `18x + 6 = 3^2`

=> `18x + 6 = 9`

=> `18x = 3`

=> `x = 1/6`

The solution of the equation `log_3 6 + log_3 (3x+1) = 2` is `x = 1/6`

rachellopez | Student, Grade 12 | (Level 1) Valedictorian

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When you add log(a) + log(b), you can rewrite this to be log(a*b). So the equation log3(6) + log3(3x+1) = 2 can be rewritten as log3(6)*(3x+1) = 2.

When you multiply the 6 by (3x+1) you have to multiply 6 by each part (distribute), which gives you log3(18x+6) = 2. You can turn the answer (2) into an exponent which gives you 18x+6 = 3^2. 3 squared equals 9. This is simple algebra from here.

18x+6 = 9

18x = 3

x = 3/18

x= 1/6

steamgirl | Student, College Junior | (Level 1) Honors

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Using the rules of logs, we know that

log3(6) + log3(3x+1) =2 is the same as

log3 (6 * (3x+1)) = 2

which then becomes the following when simplified:

log3 (18x +6) =2

Then you can use the basic definition of logs which is:

logb(m) = N is the same as b^N =m

which then gives us:

3^2 = 18x+6

simplify:

x = 1/6

Sources:

ayl0124 | Student, Grade 12 | (Level 1) Valedictorian

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First thing you must remember is that `log_3 6`  is just a number that you can input into your calculator. ``

`log_3 6 = 1.630929754...`

Solve your equation for the unknown:

`log_3 6 + log_3 (3x + 1) = 2`

`log_3 (3x+1) = 0.3690702464...`

To change a base of a log, there is a special equation:

`log_b a = (log_(10) a)/(log_(10) b)`

Therefore,

`log_3 (3x+1) = (log_(10) (3x+1))/(log_(10) 3)`

`(log (3x+1))/(log 3) = 0.3690702464...`

log 3 is just another number as well (0.477...) that you can multiply.

`log (3x +1) = 0.1760912591...`

The inverse of log is 10^.

`3x + 1 = 10^(0.1760912591...)`

`3x + 1 = 1.5`

Now, you can simply solve for "x!"

`3x = 0.5`

`x = 1/6`

taangerine | Student, Grade 12 | (Level 1) Valedictorian

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Before doing the initial problem. Know the basic log rules. In this case, we will be using the following:

1. `log_(b)m+log_(b)n=log_(b)mn`
2. `log_b(x)=y->b^(y)=x`

In order to solve the following problem:

• Using Rule #1, rewrite the problem as `log_(3)(6)(3x+1) = 2`
• Now, using rule #2, rewrite the problem as `(6)(3x+1) = 3^(2)`
• Then, simplify by distributing as solving for the exponent: `18x+6 = 9`
• What do we do next? Well, we have to isolate the variable by subtracting 6 from both sides. Leaving us with: `18x=3`
• And finally! We have to get x by itself by what? Yes, dividing 18 from both sides. Our final answer should be  `= 3/18 or 1/6`
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