# log3x - log 9 = 1

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### 5 Answers

log3x - log 9 = 1

We know that:

logx -logy = log(x/y)

log 10 =1

==> log3x -log 9= 1

==> log(3x/9) = log 10

==> 3x/9 = 10

Multiply by 9

==> 3x= 90

Divide by 3:

==>** x= 30 **

**To check:**

**log3(30)-log9 = 1**

**log90 -log9=1**

**log(90/9) = 1**

**log 10 =1**

**1= 1**

The equation log3x - log 9 = 1 has to be solved for x. Assume that the base of the logarithm is 10 as log refers to logarithm to the base 10.

For any two logarithm, the following property holds : log a - log b = log (a/b)

This gives:

log 3x - log 9 = 1

log (3x)/9 = 1

log x/3 = 1

Now, if log_b x = y, x = b^y

This gives:

x/3 = 10^1 = 10

x = 3*10

x = 30

The solution of the equation log3x - log 9 = 1 is x = 30

First, we'll discuss the constraints of existence of logarithms:

3x>0

We'll divide by 3:

x>0

So, for the logarithm to exist, the values of x have to be in the interval (0,+inf.).

Now, we'll solve the equation, using the quotient property of logarithms: the difference of logarithms is the logarithm of the quotient.

log3x - log 9 = log (3x/9) = log (x/3)

The equation will become:

log (x/3) = 1

But 1 = log 10

We'll re-write the equation:

log (x/3) = log 10

We'll use the one to one property:

(x/3) = 10

We'll multiply by 3 both sides:

x = 30

**Since the value of x is in the interval of convenient values, the equation will have the solution, namely x = 30.**

We know:

log A - log B = log (A/B)

We also know:

log 10 = 1 and

log 100 = 3

Therefore:

log 3x - log 9 = 1

==> log (3x/9) = 1 = log 10

Therefore:

3x/9 = 10

x/3 = 10

x = 10*3 = 30

Answer:

x = 30

We use the property loga-log g = log(a/g).

So we can rewrite log3x-log9 = log(3x/9). The given equation now becomes:

log(3x/9) = 1 = log10. Taking antilogarithms, we get:

3x/9 = 10. Multiply by 9.

3x= 90.

3x/3 = 90/3

x =30