Evaluate: log_{10}(x+3)=1-log_{10}(x)

Expression: $\log_{ 10 }({ x+3 })=1-\log_{ 10 }({ x })$

Determine the defined range

$\begin{array} { l }\log_{ 10 }({ x+3 })=1-\log_{ 10 }({ x }),& x \in \langle0, +\infty\rangle\end{array}$

Move the expression to the left-hand side and change its sign

$\log_{ 10 }({ x+3 })+\log_{ 10 }({ x })=1$

Use $\log_{ a }({ x })+\log_{ a }({ y })=\log_{ a }({ x \times y })$ to simplify the expression

$\log_{ 10 }({ \left( x+3 \right) \times x })=1$

Distribute $x$ through the parentheses

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

Convert the logarithm into exponential form using the fact that $\log_{ a }({ x })=b$ is equal to $x={a}^{b}$

${x}^{2}+3x={10}^{1}$

Any expression raised to the power of $1$ equals itself

${x}^{2}+3x=10$

Move the constant to the left-hand side and change its sign

${x}^{2}+3x-10=0$

Write $3x$ as a difference

${x}^{2}+5x-2x-10=0$

Factor out $x$ from the expression

$x \times \left( x+5 \right)-2x-10=0$

Factor out $-2$ from the expression

$x \times \left( x+5 \right)-2\left( x+5 \right)=0$

Factor out $x+5$ from the expression

$\left( x+5 \right) \times \left( x-2 \right)=0$

When the product of factors equals $0$, at least one factor is $0$

$\begin{array} { l }x+5=0,\\x-2=0\end{array}$

Solve the equation for $x$

$\begin{array} { l }x=-5,\\x-2=0\end{array}$

Solve the equation for $x$

$\begin{array} { l }\begin{array} { l }x=-5,\\x=2\end{array},& x \in \langle0, +\infty\rangle\end{array}$

Check if the solution is in the defined range

$x=2$