What are the answer choices for these natural log and composition problems?

Click For Summary
SUMMARY

The discussion focuses on solving natural logarithm and composition problems involving inverse functions. The first problem involves the equation x + 14 = 2g^(-1)(x) and results in g^(-1)(x) = (x + 14) / 2, which does not match any provided answer choices. The second problem, 6 = 7e^x + e^(-x), is transformed into a quadratic equation 7t² - 6t + 1 = 0, yielding solutions for x as ln[(3 ± √2)/7], which also do not correspond to the answer choices. Participants confirm the correctness of the solutions despite the discrepancies with the answer options.

PREREQUISITES
  • Understanding of natural logarithms and their properties
  • Familiarity with inverse functions and their notation
  • Basic knowledge of quadratic equations and their solutions
  • Proficiency in manipulating exponential equations
NEXT STEPS
  • Study the properties of inverse functions in detail
  • Learn how to solve quadratic equations using the quadratic formula
  • Explore the relationship between exponential and logarithmic functions
  • Practice solving complex logarithmic equations with multiple variables
USEFUL FOR

Students studying calculus, mathematics educators, and anyone looking to deepen their understanding of logarithmic and exponential functions.

geegeet
Messages
3
Reaction score
0

Homework Statement


I keep getting what I think is the "right answer", but it's not one of my choices :^(
Here are the problems and answer choices.
1).
precal1.jpg

2).
precal2.jpg


Homework Equations



nothing really...or at least I don't think besides the fact that e^-x = 1/e^x

The Attempt at a Solution



1).
x + 14 = 2 g^( - 1) ( x )
(x + 14) / 2 = g^( - 1) ( x)

x - 3 = 4 f^( - 1) ( x)
( x - 3) / 4 = f^( - 1) ( x)

g^( - 1) o f^( - 1) ( x) = (x + 53) / 8
I get this for number 1, but it is not one of the answer choices!

2).
6 = 7e^x + e^(-x )
Let e^x = t, (t > 0), thus e^(-x) = 1/t
6 = 7t + 1/t
Multiplying each side by t
6t = 7t² + 1
7t² - 6t + 1 = 0
t = (3 ± √2)/7
e^x = t ==> x = ln (t)
x = ln [(3 + √2)/7] or x = ln [(3 − √2)/7]
I get this for number 2, but it is also not one of the answer choices!
 
Physics news on Phys.org
I agree with both your answers.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
7
Views
2K
Replies
8
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
2
Views
2K