How to solve an exponential equation

  • Thread starter Thread starter Mr Davis 97
  • Start date Start date
  • Tags Tags
    Exponential
AI Thread Summary
To solve the exponential equation 8100(7/6)^{x} = 11,000, the correct approach involves taking the natural logarithm of both sides after dividing by 8100. The initial incorrect calculation involved using log base 9450, which does not align with the original equation. The correct simplification leads to (7/6)^{x} = 11000/8100, or 110/81. This method clarifies the relationship between the terms and avoids confusion with incorrect bases. The final solution for x is approximately 2.
Mr Davis 97
Messages
1,461
Reaction score
44

Homework Statement



Solve for x: 8100(\frac{7}{6})^{x} = 11,000


2. The attempt at a solution

x = log_{9450}(11,000) = 1.02

I know this is not the right answer, but I am not sure what I am doing wrong.
 
Physics news on Phys.org
Where did the 9450 come from?

EDIT: Never mind. I figured it out.

Just take the natural log of both sides:
$$\ln \left[8100\left(\frac 76\right)^x\right] = \ln 11000.$$ What do you get when you use the properties of logarithms to expand the lefthand side?

(You could divide both sides by 8100 first, and then take the log of both sides.)
 
vela said:
Where did the 9450 come from?

EDIT: Never mind. I figured it out.

Just take the natural log of both sides:
$$\ln \left[8100\left(\frac 76\right)^x\right] = \ln 11000.$$ What do you get when you use the properties of logarithms to expand the lefthand side?

(You could divide both sides by 8100 first, and then take the log of both sides.)

Thank you, I solved it. the answer is approximately 2. However, why didn't my previous attempt work? What was wrong about it?
 
Because ##x = \log_{9450} 11000## corresponds to the exponential equation ##9450^x = 11000##, but ##9450^x = \left(8100\times \frac 76\right)^x = 8100^x \left(\frac 76\right)^x##. The last expression, though similar, doesn't match what you started with on the lefthand side.
 
  • Like
Likes 1 person
It would be simpler, and less error prone, to first divide both sides by 8100:
\left(\frac{7}{6}\right)^x= \frac{11000}{8100}= \frac{110}{81}

Now take the logarithm of both sides.
 
  • Like
Likes 1 person
8100\left(\frac{7}{6}\right)^x
is NOT the same as
\left(8100*\frac{7}{6}\right)^x
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top