Exponential Equation Help with Log Tables

[cbarker1]

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

 

[earboth]

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:

\(\displaystyle -2 + 0.15806 \approx -1.84194\)

and

\(\displaystyle \log(1.84194) = 0.26528\)

This error occurs in your following calculations again.

The best would be if you keep mantissae and prefixes separated.

 

[earboth]

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

\(\displaystyle |x| = \frac{1.84194}{0.46}\)

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

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