what is the solution to the equation below? 3log₄x = log₄32 + log₄2\nx = -8\nx = -4\nx = 4\nx = 8
what is the solution to the equation below? 3log₄x = log₄32 + log₄2\nx = -8\nx = -4\nx = 4\nx = 8
Answer
Explanation:
Step1: Use log addition rule
By the rule $\log_a M+\log_a N = \log_a(MN)$, the right - hand side of the equation $3\log_4 x=\log_4 32+\log_4 2$ becomes $\log_4(32\times2)=\log_4 64$. So the equation is $3\log_4 x=\log_4 64$.
Step2: Use log power rule
By the rule $n\log_a M=\log_a M^n$, the left - hand side $3\log_4 x$ can be written as $\log_4 x^3$. So the equation is $\log_4 x^3=\log_4 64$.
Step3: Solve for x
Since $\log_a M=\log_a N$ implies $M = N$ for $a>0,a\neq1$, we have $x^3 = 64$. Taking the cube - root of both sides, $\sqrt[3]{x^3}=\sqrt[3]{64}$, so $x = 4$.
Answer:
$x = 4$