Solve Exponential Variable Equations with Logarithms

  • Thread starter Thread starter Mentallic
  • Start date Start date
  • Tags Tags
    Exponents Variable
Click For Summary
SUMMARY

The discussion centers on solving the exponential variable equation \(5^{3x} - 12^x + 2^{\frac{x}{2}} = 5\). The user attempted to apply logarithmic rules but found that taking the logarithm complicated the problem further. The approximate solution provided is \(x \approx 0.386744\), indicating that numerical approximation methods are necessary for finding solutions to this equation, as exact solutions using logarithms are not feasible.

PREREQUISITES
  • Understanding of exponential equations
  • Familiarity with logarithmic properties
  • Basic numerical approximation techniques
  • Knowledge of solving equations involving multiple bases
NEXT STEPS
  • Research numerical methods for solving nonlinear equations
  • Learn about the Newton-Raphson method for root finding
  • Study the properties of logarithms in depth
  • Explore software tools like Python's SciPy for numerical solutions
USEFUL FOR

Students, mathematicians, and educators seeking to understand the complexities of solving exponential equations and those interested in numerical approximation techniques.

Mentallic
Homework Helper
Messages
3,802
Reaction score
95

Homework Statement


Solve for x
[tex]5^{3x}-12^x+2^{\frac{x}{2}}=5[/tex]


Homework Equations


logarithm rules?
log(a+b)=?


The Attempt at a Solution


Taking the logarithm of both sides only makes things worse from what I can see.
So really, I don't know how to start.
[tex]x\approx 0.386744[/tex]
 
Physics news on Phys.org
Mentallic said:
Solve for x
[tex]5^{3x}-12^x+2^{\frac{x}{2}}=5[/tex]

hmm … that's 125x - 12x + (√2)x = 5 …

i can't see any way of solving that except by numerical approximation :redface:
 
Is this possibly due to the lack of understanding with this maths or because it cannot be solved with an exact answer? Such as 'exact' answers with a series of logarithms, even though the logs themselves are approximated.
Any idea how I could go about obtaining a reasonably accurate numerical approximation?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
3K