Logarithms

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ryan_kalle

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Working with logarithms. I have a lot of trouble dealing with fractions and exponents.


Find the exact solution
log base x(1/9)=-2/3
x^(-2/3)=1/9
I lose it there

Find the approximate solution to each equation. Round to 4 places
1/(10^x)=2
No clue

Solve for t
20=10e^(.02xt)

Rewrite each expression in terms of log base a(5)
log base a (1/5)
No clue
 
\(\displaystyle log_x\frac{1}{9} \ = \ \frac{-2}{3}\)

\(\displaystyle \frac{1}{9} \ = \ x^{-2/3}\)

\(\displaystyle \frac{1}{9} \ = \ \frac{1}{x^{2/3}}\)

\(\displaystyle x^{2/3} \ = \ 9\)

\(\displaystyle x \ = \ 9^{3/2} \ = \ 27\)
 
From the Denis Property of Logarithms, we have the following.

x = (1/9)^(-3/2)

One step leads to:

x = 9^(3/2) = 9 ?9

I would have done the exercise as Glenn did. Neat shortcut, Denis.
 
ryan_kalle said:
1/(10^x)=2
No clue

Solve for t
20=10e^(.02xt)

Rewrite each expression in terms of log base a(5)
log base a (1/5)
No clue
Click to expand...

You really need to know what logarithms are and why we use them, before understanding exercises like these. You also need to study the properties of exponents and logarithms, until you can use them as tools, before understanding how to solve exercises like these. Are you taking a class?

Each of the steps below comes from using a property.

Given: 1/(10^x) = 2

10^(-x) = 2

-x log(10) = log(2)

Solve this last equation for x, and round the decimal approximation to four places.

--------------------------------------------

20=10e^(.02xt)

[I guess that symbol x is a variable.]

Isolate the power of e, before taking logarithms, and simplify the rational number that results.

e^(0.02 x t) = 2

Use natural logarithms, of course, and then apply the property that tells us how to get the exponent "down in front".

0.02 x t ln(e) = ln(2)

Solve for t, and simplify (rounding to four places).

--------------------------------------------

Rewrite log[base a](1/5) in terms of log[base a](5)

log[base a](1/5) = -log[base a](5)

See if you can figure out why.

I welcome specific questions about these steps. 8-)
 
Hello, ryan_kalle!

\(\displaystyle \text{Find the exact solution: }\;\log_x\left(\frac{1}{9}\right) \:=\:-\frac{2}{3}\)
Click to expand...

\(\displaystyle \text{Rewrite the equation: }\;x^{-\frac{2}{3}} \:=\:\frac{1}{9} \;=\;\frac{1}{3^2} \;=\;3^{-2}\)

\(\displaystyle \text{Raise to the }-\tfrac{3}{2}\text{ power: }\;\left(x^{-\frac{2}{3}}\right)^{-\frac{3}{2}} \;=\;\left(3^{-2}\right)^{-\frac{3}{2}}\)

\(\displaystyle \text{And we have: }\;x \;=\;3^3 \quad\Rightarrow\quad x \:=\:27\)



\(\displaystyle \text{Find the approximate solution. }\:\text{Round to 4 places. }\;\;\frac{1}{10^x} \:=\:2\)
Click to expand...

\(\displaystyle \text{We have: }\;10^{-x} \;=\;2\)

\(\displaystyle \text{Take logs, base 10: }\;\log\left(10^{-x}\right) \;=\;\log(2)\)

. . \(\displaystyle -x\underbrace{\log(10)}_{\text{This is 1}} \;=\;\log(2) \quad\Rightarrow\quad -x \;=\;\log(2)\)

\(\displaystyle \text{Therefore: }\;x \;=\;-\log(2) \;\approx\;-0.3010\)



\(\displaystyle \text{Solve for }t\!:\;\;10e^{0.02t} \:=\:20\)
Click to expand...

\(\displaystyle \text{Divide by 10: }\;e^{0.02t} \;=\;2\)

\(\displaystyle \text{Take logs, base }e\!:\;\;\ln\left(e^{0.02t}\right) \;=\;\ln(2)\)

. . \(\displaystyle 0.02t\!\!\underbrace{\ln(e)}_{\text{This is 1}} \;=\;\ln(2) \quad\Rightarrow\quad 0.02t \;=\;\ln(2)\)

\(\displaystyle \text{Therefore: }\;t \;=\;\frac{\ln(2)}{0.02} \;\approx\;34.657\)



\(\displaystyle \text{Rewrite the expression in terms of }\log_a(5)\)

. . \(\displaystyle \log_a\left(\frac{1}{5}\right)\)
Click to expand...

\(\displaystyle \log_a\left(\frac{1}{5}\right) \;=\;\log_a\left(5^{-1}\right) \;=\;-\log_a(5)\)

 
Thank you all for your help. Because of you assistance I was able to ace my latest test.
 
well im working with logarithms and i could really usse someones help right now. well my question goes like this:
log x= -1.75
im really confused and dont know what to do?
i tried to do : Ln e^(-1.75)= x
but its wrong.
 
Hi yv3tt3692:

In the future, please use the [newtopic] button on the index page, to start your own thread.

This is an easy exercise, once we understand the definition of "logarithm".

The notation "log" means base-10, unless specified otherwise.

In other words, the symbolism log(x) represents some exponent on 10.

We're told that log(x) is the number -1.75; this means the exponent on the base is -1.75.

For example: if they had written log(x) = 4.5, instead, then we would be talking about 10^4.5, instead.

Do you realize that the variable x represents the value of a power of 10 ?

In my example, log(x) = 4.5, the value of the power of 10 is 10^4.5 = 31622.7766 (rounded from calculator result).

Clearly, then, we have x = 31622.7766 because we see that log(31622.7766) = 4.5

Again, logarithms are exponents. The line above shows that 4.5 is a logarithm, so 4.5 must be an exponent.

log means base-10, so 4.5 is the exponent on 10.

log(x) = 4.5 This is called the "logarithmic form" of the relationship in my example

x = 10^4.5 This is called the "exponential form" of the same relationship

These two forms are equivalent statements about the relationship between 10, 4.5, and x. We need to be able to switch back and forth between these two forms, when working with logarithms.

I hope this short explanation makes sense to you, and that you now know how to find the value of x in your exercise.

Since the given logarithm equals -1.75, that means -1.75 is an exponent, and I already told you what the base is. Use a calculator, to find the power of 10 which is x. 8-)

If you're still confused, please tell me why.

Cheers ~ Mark
 
\(\displaystyle What \ I \ went \ through \ in \ High \ School.\)

\(\displaystyle Solve \ for \ x: \ log_{10}x \ = \ -1.75\)

\(\displaystyle log_{10}x \ = \ -1.75\)

\(\displaystyle log_{10}x \ = \ 8.25-10\)

\(\displaystyle Now, \ pulling \ out \ a \ book \ of \ logs \ to \ four \ \ places, \ we \ find \ (corresponding \ to \ the \ mantissa \ 2504)\)

\(\displaystyle the \ number \ 178. \ Hence, \ since \ the \ characteristic \ is \ 8-10 \ = \ -2, \ the \ required \ number \ is\)

\(\displaystyle \dot= \ .0178, \ note \ I \ didn't \ interpolate \ (a \ real \ drudge).\)

\(\displaystyle Nowadays, \ with \ my \ trusty \ TI-89, \ I \ just \ plug \ in \ 10^{-1.75} \ and \ get \ .0177827941.\)
 
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