Question

1231c2_25-26 - c - rusd math 3 semester b_90-38578-9344-1231c2
solving equations using properties of logarithms
solving logarithmic equations using the def
solve the equation \\(\log_{6}(13 - x) = 1\\).
\\(x = \square\\)

Explanation:

Step1: Recall the definition of logarithm

The logarithmic equation $\log_{a}b = c$ is equivalent to the exponential equation $a^{c}=b$ (where $a>0,a
eq1$ and $b > 0$). For the equation $\log_{6}(13 - x)=1$, we can use this definition. Here, $a = 6$, $c=1$ and $b=13 - x$. So we can rewrite the logarithmic equation as an exponential equation: $6^{1}=13 - x$.

Step2: Solve the resulting linear equation

We know that $6^{1}=6$, so the equation becomes $6=13 - x$. To solve for $x$, we can add $x$ to both sides of the equation: $6 + x=13$. Then subtract 6 from both sides: $x=13 - 6$.

Step3: Calculate the value of x

Calculating $13 - 6$, we get $x = 7$. We should also check if the solution is valid in the original logarithmic equation. Substitute $x = 7$ into the argument of the logarithm: $13-7 = 6>0$, so the solution is valid.

Answer:

$x = 7$