Solve for x log of 5x+ log of 2x=1

$\log (5 x) + \log (2 x) = 1$

Step 1

Simplify the left side.

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

$\log (5 x (2 x)) = 1$

Rewrite using the commutative property of multiplication.

$\log (5 \cdot (2 x \cdot x)) = 1$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$\log (5 \cdot (2 (x \cdot x))) = 1$

Multiply $x$ by $x$ .

$\log (5 \cdot (2 x^{2})) = 1$

Multiply $5$ by $2$ .

$\log (10 x^{2}) = 1$

Step 2

Rewrite $\log (10 x^{2}) = 1$ 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 $b^{y} = x$ .

$10^{1} = 10 x^{2}$

Step 3

Solve for $x$ .

Rewrite the equation as $10 x^{2} = 10^{1}$ .

$10 x^{2} = 10$

Divide each term in $10 x^{2} = 10^{1}$ by $10$ and simplify.

Divide each term in $10 x^{2} = 10^{1}$ by $10$ .

$\frac{10 x^{2}}{10} = \frac{10^{1}}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 x^{2}}{10} = \frac{10^{1}}{10}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{10^{1}}{10}$

Simplify the right side.

Cancel the common factor of $10^{1}$ and $10$ .

Factor $10$ out of $10^{1}$ .

$x^{2} = \frac{10 \cdot 1}{10}$

Cancel the common factors.

Factor $10$ out of $10$ .

$x^{2} = \frac{10 \cdot 1}{10 (1)}$

Cancel the common factor.

$x^{2} = \frac{10 \cdot 1}{10 \cdot 1}$

Rewrite the expression.

$x^{2} = \frac{1}{1}$

Rewrite the expression.

$x^{2} = 1$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$x = \pm 1$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 1$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 1$

Step 4

Exclude the solutions that do not make $\log (5 x) + \log (2 x) = 1$ true.

$x = 1$

Solve for x log of 5x+ log of 2x=1

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