Answer:
Your answer is correct.
[tex]\sf w = \dfrac{e^5 - 9}{4} \approx 34.85 [/tex]
Step-by-step explanation:
To solve the equation [tex]\sf \ln(4w + 9) = 5 [/tex], we can use the property of logarithms that states [tex]\sf \ln(a) = b [/tex] is equivalent to [tex]\sf a = e^b [/tex].
Given:
[tex]\sf \ln(4w + 9) = 5 [/tex]
By applying the property of logarithms, we can exponentiate both sides of the equation to eliminate the natural logarithm:
[tex]\sf 4w + 9 = e^5 [/tex]
Next, solve for [tex]\sf w [/tex]:
Subtract 9 from both sides:
[tex]\sf 4w = e^5 - 9 [/tex]
Divide both sides by 4:
[tex]\sf w = \dfrac{e^5 - 9}{4} [/tex]
Now, we need to check if there are any extraneous solutions. Since the natural logarithm function [tex]\sf\ln(x)[/tex] is only defined for [tex]\sf x > 0[/tex], the expression inside the logarithm must be positive:
[tex]\sf 4w + 9 > 0 [/tex]
Solving for [tex]\sf w [/tex]:
[tex]\sf 4w > -9 [/tex]
[tex]\sf w > -\dfrac{9}{4} [/tex]
Given that [tex]\sf e^5 [/tex] is a positive number (approximately 148.413), [tex]\sf e^5 - 9 [/tex] is also positive. Thus, [tex]\sf w = \dfrac{e^5 - 9}{4} [/tex] is greater than [tex]\sf -\dfrac{9}{4} [/tex], and there are no extraneous solutions.
Therefore, the solution to the equation is:
[tex]\sf w = \dfrac{e^5 - 9}{4} [/tex]
This can be approximated using the value of [tex]\sf e^5 \approx 148.4131591 [/tex]:
[tex]\sf w \approx \dfrac{148.4131591 - 9}{4} \\\\ \approx \dfrac{139.4131591}{4} \\\\\approx 34.85328978\\\\ \approx 34 .85 \textsf{(in 2 d.p.)} [/tex]
So, the solution is:
[tex]\sf w = \dfrac{e^5 - 9}{4} \approx 34.85 [/tex]
Note:
[tex] \boxed{\boxed{\begin{aligned} &\sf \bold{\underline{\textsf{Some rule of Logorithm}}} \\&\textsf{Product Rule:} && \sf \log_b(xy) = \log_b(x) + \log_b(y) \\&\textsf{Quotient Rule:}&&\sf \log_b\left(\dfrac{x}{y}\right) = \log_b(x) - \log_b(y) \\& \textsf{Power Rule:} && \sf \log_b(x^r) = r \log_b(x) \\& \textsf{Change of Base Formula:} && \sf \log_b(x) = \dfrac{\log_c(x)}{\log_c(b)} (\textsf{for any base } c) \\& \textsf{Logarithm of 1:} && \sf \log_b(1) = 0 \\& \textsf{Logarithm of a Base:} && \sf \log_b(b) = 1 \\& \textsf{Negative Logarithm:}&&\sf \log_b\left(\dfrac{1}{x}\right) =-\log_b(x)\\& \textsf{Sum Rule:} && \sf \log_b(x \cdot y)= \log_b(x) + \log_b(y) \\& \textsf{Difference Rule:} &&\sf \log_b\left(\dfrac{x}{y}\right) = \log_b(x) -\log_b(y)\end{aligned}}}[/tex]
