Solve Exponential: Division w/ln - Answer Not Found

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In summary: I don't have a TI-30X IIS, but I think you need to enter it like this:-2*ln(5) / (7*ln(5) - ln(3))I don't have a TI-30X IIS, but I think you need to enter it like this:-2*ln(5) / (7*ln(5) - ln(3))Yes, that is what I did and got the same answer as you. But I put the parentheses a bit differently and got -0.322 or so. 0.005 off. I am trying to figure out why. I cannot get the answer goosey00 posted.
  • #1
goosey00
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How do you divide using ln:

my example is x(7ln5-ln3)=-2ln5 and divide 7ln5-ln3 by both side to get x. So I have -2ln5/7ln5-ln3 and I can't seem to find the answer.
 
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  • #2
goosey00 said:
How do you divide using ln:

my example is x(7ln5-ln3)=-2ln5 and divide 7ln5-ln3 by both side to get x. So I have -2ln5/7ln5-ln3 and I can't seem to find the answer.

Hi goosey00,

Try posting as much information as you can each time. This will save time by not having as many followup questions. :)

If you begin with \(\displaystyle x \left( 7 \ln(5)- \ln(3) \right) = -2 \ln(5)\) then indeed one form of the answer is \(\displaystyle x = \frac{-2 \ln(5)}{\left( 7 \ln(5)- \ln(3) \right)}\). Since this is multiple choice though, they want you to use the rules I mentioned to you yesterday about logarithms to simplify this answer further.

Another way to think of those two rules is (1) an exponent inside the natural log can be put in front of the whole expression and the reverse, (2)the natural log of a fraction is the natural log of the top minus the natural log of the bottom, and the reverse.

Remember that \(\displaystyle a \ln(x) = \ln(x^a)\) and \(\displaystyle \ln(a/b)=\ln(a)-\ln(b)\). Using those two rules you should be able to simplify things.
 
  • #3
Its not multiple choice, they want me to divide and when I enter it in my calculator, I must be entering it wrong. I am entering ln(-2)5/ln(7)5-ln(3). Please don't laugh, I just am clueless. (Talking)
 
  • #4
You have found:

$\displaystyle x=-\frac{2\ln(5)}{7\ln(5)-\ln(3)}$

or

$\displaystyle x=\frac{2\ln(5)}{\ln(3)-7\ln(5)}$

You may use a calculator to get a decimal approximation for this value. IN the days of log tables. this would probably be the preferred form. If you are interested in simplifying the result, you may proceed as follows:

Divide each term by $\displaystyle \ln(5)$ to get:

$\displaystyle x=\frac{2}{\frac{\ln(3)}{\ln(5)}-7}$

This is perhaps simpler to use a calculator for which to get a decimal approximation. We may make one more change if we wish, but then most calculators will not give an approximation for this form:

Using the change of base formula, we know:

$\displaystyle \frac{\ln(3)}{\ln(5)}=\log_5(3)$ hence:

$\displaystyle x=\frac{2}{\log_5(3)-7}$

edit: Oops, sorry to post after help is given! :)
 
  • #5
goosey00 said:
Its not multiple choice, they want me to divide and when I enter it in my calculator, I must be entering it wrong. I am entering ln(-2)5/ln(7)5-ln(3). Please don't laugh, I just am clueless. (Talking)

ln(-2) is impossible first of all :) The log and natural log aren't defined for 0 or negative numbers. This also depends on what kind of calculator you have. Hopefully you can enter in long expressions before hitting "equals". If so you should type something like:

(-2*ln(5))/(7*ln(5)-ln(3))

Be careful with parentheses. When you open one you must always close it somewhere. When I do it on our site's calculator (located under "MHB Widgets") I get -.317 or so.
 
  • #6
I couldn't do the long way on the calculator. I think I can just break it down but to get your answer I was off 2 decimal places. Is that normal?
 
  • #7
goosey00 said:
I couldn't do the long way on the calculator. I think I can just break it down but to get your answer I was off 2 decimal places. Is that normal?

What do you mean off by 2 decimal places? It was plus or minus .01 from my answer? What kind of calculator do you have again?
 
  • #8
Ti30x
 
  • #9
I did it by doing each separately and got it as I then divided it both out. So thanks, it worked..
 
  • #11
My TI-89 returns ≈ -0.317
 

FAQ: Solve Exponential: Division w/ln - Answer Not Found

What is an exponential equation?

An exponential equation is an equation in the form of y = ab^x, where a and b are constants and x is the variable. Exponential equations often involve exponential functions, which are functions in which the variable is the exponent.

How do you solve an exponential equation?

To solve an exponential equation, you can use the properties of logarithms, specifically the Product and Quotient Rules. You can also take the natural logarithm (ln) of both sides to convert the equation into a linear form, which can then be solved using algebraic methods.

What is the rule for division with logarithms?

The rule for division with logarithms is to subtract the logs of the two numbers. This can be written as log(a/b) = log(a) - log(b). This rule can be used to simplify complex logarithmic expressions and solve exponential equations.

Why is the answer not found when solving exponential equations with division and logarithms?

This can happen when the equation has no real solution. In other words, there is no value of the variable that will make the equation true. This can occur when taking the logarithm of a negative number or when the base of the exponential function is not greater than 1.

Can you check your answer when solving exponential equations with division and logarithms?

Yes, you can always check your answer by plugging it back into the original equation and simplifying. If the resulting equation is true, then your answer is correct. This is an important step to ensure accuracy when solving exponential equations.

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