Question

solve the following equations. if necessary, round your final answer to four decimal places.
a) \\(\log_{3}(4x + 11) = 4\\)
b) \\(3^{x} = 29\\)

Explanation:

Part (a)

Step1: Convert log to exponential

Using the definition of logarithms: if $\log_b(a) = c$, then $b^c = a$. So for $\log_3(4x + 11) = 4$, we get $3^4 = 4x + 11$.
$3^4 = 81$, so the equation becomes $81 = 4x + 11$.

Step2: Solve for x

Subtract 11 from both sides: $81 - 11 = 4x$.
$70 = 4x$.
Divide both sides by 4: $x = \frac{70}{4} = 17.5$.

Step1: Take log of both sides

Take the natural logarithm (ln) of both sides of $3^x = 29$: $\ln(3^x) = \ln(29)$.
Using the logarithm power rule $\ln(a^b) = b\ln(a)$, we get $x\ln(3) = \ln(29)$.

Step2: Solve for x

Divide both sides by $\ln(3)$: $x = \frac{\ln(29)}{\ln(3)}$.
Calculate $\ln(29) \approx 3.3672958248682004$ and $\ln(3) \approx 1.0986122886681132$.
Then $x \approx \frac{3.3672958248682004}{1.0986122886681132} \approx 3.0616$.

Answer:

$x = 17.5$

Part (b)