Exponential Equation Help with Log Tables

  • Context: MHB 
  • Thread starter Thread starter cbarker1
  • Start date Start date
  • Tags Tags
    Exponential Log
Click For Summary
SUMMARY

The discussion focuses on solving the exponential equation ${2.884}^{x}=0.01439$ using logarithmic tables. The correct method involves calculating $x$ as $x=\frac{\log\left({0.01439}\right)}{\log\left({2.884}\right)}$, leading to the solution of $x \approx -4.004$. Participants identified errors in logarithmic calculations, specifically in evaluating $\log\left({0.01439}\right)$ and $\log\left({2.884}\right)$. The correct evaluation of these logarithms is crucial for accurate results.

PREREQUISITES
  • Understanding of exponential equations
  • Familiarity with logarithmic properties
  • Ability to use logarithm tables
  • Basic arithmetic operations
NEXT STEPS
  • Learn how to accurately use logarithm tables for calculations
  • Study the properties of logarithms, including change of base formula
  • Practice solving exponential equations with various bases
  • Explore the historical context and applications of logarithms in mathematics
USEFUL FOR

Students, educators, and anyone interested in mastering logarithmic calculations and solving exponential equations using traditional methods.

cbarker1
Gold Member
MHB
Messages
345
Reaction score
23
Directions: Use a log table to solve for x:

${2.884}^{x}=0.01439$

$x*\log\left({2.884}\right)=\log\left({0.01439}\right)$

$x=\frac{\log\left({0.01439}\right)}{\log\left({2.884}\right)}$ is the exact answer.

The solution to the problem is -4.004 in the back of the book.

To evaluate the logarithms with table:

$\log\left({.01439}\right)\equiv\log\left({1.439}\right)-2, where \log\left({1.439}\right)=.15806$

$-2.15806, 8.15806-10$

$\log\left({2.884}\right)=.46000$

$x=\frac{-2.15806}{.46000}$ drop the negative sign to compute the logarithms.

$\log\left({\frac{2.15806}{.46}}\right)=\log\left({2.15806}\right)-\log\left({.46}\right)$

$\log\left({2.15806}\right)=.3340512$

$.3340512, 10.3340512-10$

$\log\left({.4600}\right)\equiv\log\left({4.600}\right)-1, where \log\left({4.600}\right)=.66276$

$-1.66276, 9.66276-10$

Now, I need some help to subtract the correct values of $\log\left({2.15806}\right)$ and $\log\left({.46000}\right)$ to get the answer of .60249 in the log table.Thanks for the help

CBarker1
 
Last edited:
Physics news on Phys.org
Cbarker1 said:
Directions: Use a log table to solve for x:

${2.884}^{x}=0.01439$

$x*\log\left({2.884}\right)=\log\left({0.01439}\right)$

$x=\frac{\log\left({0.01439}\right)}{\log\left({2.884}\right)}$ is the exact answer.

The solution to the problem is -4.004 in the back of the book.

To evaluate the logarithms with table:

$\log\left({.01439}\right)\equiv\color{red}\log\left({1.439}\right)-2, where \log\left({1.439}\right)=.15806$

$\color{red}-2.15806, 8.15806-10$

...

Good morning,

I've marked in red the calculations where you made a mistake:

$$-2 + 0.15806 \approx -1.84194$$

and

$$\log(1.84194) = 0.26528$$

This error occurs in your following calculations again.

The best would be if you keep mantissae and prefixes separated.
 
Cbarker1 said:
Directions: Use a log table to solve for x:

...

Hello again,

I'll show you how I've learned to use a log table. (I visited school without calculators or computers. The most advanced piece of technology was a slide-ruler!)

You want to calculate

$$|x| = \frac{1.84194}{0.46}$$

with a log table. "op" means operation of the logarithms, N is the numerus and log means the logarithm base 10.

$$\begin{array}{c|l|c|l}op & N & & log \\ \hline \text{-} & 1.84194 & \rightarrow & 0.26528 \\ & 0.46 & \rightarrow & 0.66276 - 1 \\ \hline & 4.0042 & \leftarrow & 0.60252 \end{array}$$
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
3
Views
2K
Replies
17
Views
3K