Solve for z log of z+ log of 2+3=1

$\log (z) + \log (2 + 3) = 1$

Simplify the left side.

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

$\log (z (2 + 3)) = 1$

Add $2$ and $3$ .

$\log (z \cdot 5) = 1$

Move $5$ to the left of $z$ .

$\log (5 z) = 1$

Rewrite $\log (5 z) = 1$ 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 $b^{y} = x$ .

$10^{1} = 5 z$

Solve for $z$ .

Rewrite the equation as $5 z = 10^{1}$ .

$5 z = 10$

Divide each term in $5 z = 10^{1}$ by $5$ and simplify.

Divide each term in $5 z = 10^{1}$ by $5$ .

$\frac{5 z}{5} = \frac{10^{1}}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 z}{5} = \frac{10^{1}}{5}$

Divide $z$ by $1$ .

$z = \frac{10^{1}}{5}$

Simplify the right side.

Evaluate the exponent.

$z = \frac{10}{5}$

Divide $10$ by $5$ .

$z = 2$

Solve for z log of z+ log of 2+3=1

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