Solve for x 2 log of x=3+ log of x/10

$2 \log (x) = 3 + \log (\frac{x}{10})$

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

$2 \log (x) - \log (\frac{x}{10}) = 3$

Simplify each term.

Rewrite $\frac{x}{10}$ as $x \cdot 10^{- 1}$ .

$2 \log (x) - \log (x \cdot 10^{- 1}) = 3$

Rewrite $\log (x \cdot 10^{- 1})$ as $\log (x) + \log (10^{- 1})$ .

$2 \log (x) - (\log (x) + \log (10^{- 1})) = 3$

Use logarithm rules to move $- 1$ out of the exponent.

$2 \log (x) - (\log (x) - \log (10)) = 3$

Logarithm base $10$ of $10$ is $1$ .

$2 \log (x) - (\log (x) - 1 \cdot 1) = 3$

Multiply $- 1$ by $1$ .

$2 \log (x) - (\log (x) - 1) = 3$

Apply the distributive property.

$2 \log (x) - \log (x) + 1 = 3$

Multiply $- 1$ by $- 1$ .

$2 \log (x) - \log (x) + 1 = 3$

Subtract $\log (x)$ from $2 \log (x)$ .

$\log (x) + 1 = 3$

Move all terms not containing $\log (x)$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$\log (x) = 3 - 1$

Subtract $1$ from $3$ .

$\log (x) = 2$

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

$10^{2} = x$

Solve for $x$ .

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

$x = 10^{2}$

Raise $10$ to the power of $2$ .

$x = 100$

Solve for x 2 log of x=3+ log of x/10

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