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