# Solve for x. `2 - log_3 2 = log_3 x` .

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### 2 Answers

You need to move the term `-log_3 2` to the right, as `log_3 2` , such that:

`2 = log_3 2 + log_3 x`

You need to convert the sum `log_3 2 + log_3 x` into the logarithm of product, using logarithmic identity, such that:

`2 = log_3(2*x) => 2x = 3^2 => 2x = 9 => x = 9/2`

You need to test the value `x = 9/2` in equation, such that:

`2 = log_3 2 + log_3 (9/2)`

`2 = log_3 2 + log_3 9 - log_3 2`

Reducing duplicate terms yields:

`2 = log_3 9 => 2 = log_3 3^2 => 2 = 2*log_3 3` valid

**Hence, evaluating the solution to the given equation yields **`x = 9/2.`

For the logarithm to exist, x has to be positive.

We'll write 2 as:

2*1 = 2* log 3 3

We'll use the power property of logarithms and the symmetric property:

log 3 (x) = log 3 (3)^2 - log 3 (2)

Because the bases are matching, we'll transform the difference of logarithms from the right side, into a quotient. We'll apply the formula:

lg a - lg b = lg (a/b)

We'll substitute a by 9 and b by 2. The logarithms from formula are decimal logarithms. We notice that the base of logarithm is 3.

log 3 (x) = log 3 (9/2)

Because the bases are matching, we'll apply the one to one property:

x = 9/2

x = 4.5

Since the value of x is positive, the solution of the equation is admissible.