Question

hw 12 solving exponential and logarithmic equations due wednesday by 11:59pm points 100 submitting an external tool question find the value of $x$ which satisfies the following equation. $log_6(-24x + 168)-log_6(4)=1$ do not include \$x =$\ in your answer. provide your answer below:

Explanation:

Step1: Apply log - subtraction rule

Use the rule $\log_a M-\log_a N=\log_a\frac{M}{N}$. So, $\log_6(-24x + 168)-\log_6(4)=\log_6\frac{-24x + 168}{4}=1$.

Step2: Rewrite in exponential form

Since $\log_a b = c$ is equivalent to $a^c=b$, we have $6^1=\frac{-24x + 168}{4}$.

Step3: Solve for x

First, simplify the right - hand side equation: $6=\frac{-24x + 168}{4}$. Multiply both sides by 4: $24=-24x + 168$. Then, subtract 168 from both sides: $24-168=-24x$, which gives $-144=-24x$. Divide both sides by - 24: $x = 6$.

Answer:

6