Help with using log rules to Solve 5000*1.87^x=9000+800(X) for X

[saltyyy]

Hi can anyone help with this
Solve for X

5000*1.87^x=9000+800(X)

This may look like a very simple question but it is not.
Using a wolframaplha i got the answer to be x = ~1.08 and 11.2

I am pretty sure you are only suppose to use simple log rules to solve this question

Thx

 

[Deleted member 4993]

Hi can anyone help with this
Solve for X

5000*1.87^x=9000+800(X)

This may look like a very simple question but it is not.
Using a wolframaplha i got the answer to be

I am pretty sure you are only suppose to use simple log rules to solve this question

Thxs

1.87^x = 1.8 + 0.16 * x

This is a non-linear equation - beyond quadratic. It does not have a closed form solution.

How are you "pretty sure you are only suppose to use simple log rules to solve this question"?

 

[stapel]

I am pretty sure you are only suppose to use simple log rules to solve this question:

What, exactly, did the instructions for this exercise say?

Solve for X

5000*1.87^x=9000+800(X)

Since "x" and "X" are not the same variable, you can solve without using logs:

. . . . .\(\displaystyle 5000\, \left(1.87^x\right)\, =\, 9000\, +\, 800X\)

. . . . .\(\displaystyle 50\, \left(1.87^x\right)\, =\, 90\, +\, 8X\)

. . . . .\(\displaystyle 25\, \left(1.87^x\right)\, =\, 45\, +\, 4X\)

. . . . .\(\displaystyle 25\, \left(1.87^x\right)\, -\, 45\, =\, 4X\)

. . . . .\(\displaystyle \dfrac{25\, \left(1.87^x\right)\, -\, 45}{4}\, =\, X\)

If you meant "x" to be "X", however, the previous responses regarding the unsolve-ability remain correct.

 

[saltyyy]

What, exactly, did the instructions for this exercise say?


Since "x" and "X" are not the same variable, you can solve without using logs:

. . . . .\(\displaystyle 5000\, \left(1.87^x\right)\, =\, 9000\, +\, 800X\)

. . . . .\(\displaystyle 50\, \left(1.87^x\right)\, =\, 90\, +\, 8X\)

. . . . .\(\displaystyle 25\, \left(1.87^x\right)\, =\, 45\, +\, 4X\)

. . . . .\(\displaystyle 25\, \left(1.87^x\right)\, -\, 45\, =\, 4X\)

. . . . .\(\displaystyle \dfrac{25\, \left(1.87^x\right)\, -\, 45}{4}\, =\, X\)

If you meant "x" to be "X", however, the previous responses regarding the unsolve-ability remain correct.

Hi we only learned basic log rules so i assumed we only use basic law rules. But now i know you are suppose to use trail and error. Thxs for the help anyways.