Solve for x log of 3x+ log of 9=0

$\log (3 x) + \log (9) = 0$

Step 1

Simplify the left side.

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log (3 x \cdot 9) = 0$

Multiply $9$ by $3$ .

$\log (27 x) = 0$

Step 2

Rewrite $\log (27 x) = 0$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$10^{0} = 27 x$

Step 3

Solve for $x$ .

Rewrite the equation as $27 x = 10^{0}$ .

$27 x = 10^{0}$

Anything raised to $0$ is $1$ .

$27 x = 1$

Divide each term in $27 x = 1$ by $27$ and simplify.

Divide each term in $27 x = 1$ by $27$ .

$\frac{27 x}{27} = \frac{1}{27}$

Simplify the left side.

Cancel the common factor of $27$ .

Cancel the common factor.

$\frac{27 x}{27} = \frac{1}{27}$

Divide $x$ by $1$ .

$x = \frac{1}{27}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{27}$

Decimal Form:

$x = 0.\overline{037}$

Solve for x log of 3x+ log of 9=0

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