Question
Determine whether each $x$-value is a solution of the equation.$3 e^{x+2}=75$(a) $x=-2+e^{25}$(b) $x=-2+\ln 25$(c) $x \approx 1.2189$
Step 1
We do this by dividing both sides of the equation by 3. This gives us: \[e^{x+2} = \frac{75}{3} = 25\] Show more…
Show all steps
Your feedback will help us improve your experience
Ankit Gupta and 93 other Algebra educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Determine whether each $x$ -value is a solution (or an approximate solution) of the equation. $3 e^{x+2}=75$ (a) $x=-2+e^{25}$ (b) $x=-2+\ln 25$ (c) $x \approx 1.219$
Exponential and Logarithmic Functions
Exponential and Logarithmic Equations
Determine whether each $x$-value is a solution of the equation. $\ln (2+x)=2.5$ (a) $x=e^{2.5}-2$ (b) $x \approx \frac{4073}{400}$ (c) $x=\frac{1}{2}$
Solving Exponential and Logarithmic Equations
In Exercises 1– 8, determine whether each x-value is a solution of the equation. $$ \begin{array}{l}{3 e^{x+2}=75} \\ {\text { (a) } x=-2+e^{25}} \\ {\text { (b) } x=-2+\ln 25} \\ {\text { (c) } x \approx 1.2189}\end{array} $$
Transcript
600,000+
Students learning Algebra with Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD