Solve for x log base 20 of x=1+ log base 20 of 10

$\log_{20} (x) = 1 + \log_{20} (10)$

Step 1

Move all the terms containing a logarithm to the left side of the equation.

$\log_{20} (x) - \log_{20} (10) = 1$

Step 2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{20} (\frac{x}{10}) = 1$

Step 3

Rewrite $\log_{20} (\frac{x}{10}) = 1$ 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 $b^{y} = x$ .

$20^{1} = \frac{x}{10}$

Step 4

Solve for $x$ .

Rewrite the equation as $\frac{x}{10} = 20^{1}$ .

$\frac{x}{10} = 20$

Multiply both sides of the equation by $10$ .

$10 \frac{x}{10} = 10 \cdot 20^{1}$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$10 \frac{x}{10} = 10 \cdot 20^{1}$

Rewrite the expression.

$x = 10 \cdot 20^{1}$

Simplify the right side.

Simplify $10 \cdot 20^{1}$ .

Evaluate the exponent.

$x = 10 \cdot 20$

Multiply $10$ by $20$ .

$x = 200$

Solve for x log base 20 of x=1+ log base 20 of 10

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