# Solve for x equation 1+log3 x =2 .

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### 3 Answers

We have 1 + log(3)x = 2

Use 1 = log(3) 3 and log a + log b = log a*b

=> log(3) 3 + log(3) x = 2

=> log(3) 3*x = 2

=> 3x = 3^2

=> 3x = 9

=> x = 3

**The solution of the equation is x = 3**

To solve `1+log_3 x =2` use the definition of logarithm. `If log_b a = c` , `a = b^c`

`1+log_3 x =2`

Subtract 1 from both the sides

`log_3 x = 1`

x = 3^1 = 3

The solution of the equation `1+log_3 x =2` is x = 3

The solution of the equation has to respect the constraint of existence of logarithm:

x > 0

We'll write the value 1 = log3 3

We'll use the produt rule of logarithms:

log3 3 + log3 x = log3 (3x)

We'll solve the equation, taking anti-log:

log3 (3x) = 2

3x = 3^2

3x = 9

We'll divide by 3:

x = 3

**Since the value of x is positive, we'll accept it as solution of the equation, so x = 3.**