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

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

Step 1

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

$\log (2 x) = 1$

Step 2

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

$10^{1} = 2 x$

Step 3

Solve for $x$ .

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

$2 x = 10$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Evaluate the exponent.

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

Divide $10$ by $2$ .

$x = 5$

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

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