Some tricky exponentioal equations

[xortan]

Homework Statement

e^(4x-5)=9e^(x+5)

Homework Equations

log rules

The Attempt at a Solution

I have tried this one a few times using slightly different methods and getting 2 answers and neither of them seem to be working when i plug them back into the equation.

Here is my first method...

1. Divide out the e^(x+5) so i get e^(4x-5)/e^(x+5) = 9

2. Take the natural log of everything so i end up with

(4x-5)lne/(x+5)lne = ln 9

3. After doing the algebra i got the answer to be 8.87 (rounded to 3 sig figs), but it wasnt checking out when i plugged it back into the equation

Here is my second method...

1. I just started going crazy with the natural logs getting

(4x-5)lne - (x+5) ln 9e

The ln9e don't quite sit right with me tho, however after doing the algebra i get x=-10, this doesn't seem to check out either...please help!

Homework Statement

(3^7x)(27^x)=9

Homework Equations

logs

The Attempt at a Solution

Alright well since i know that they are all powers of 3 i changed the bases so the equation became

(3^7x)((3^3)x)=3^2

Then i just looked at the exponents and ended up with

21x^2 = 2

Did the algebra and this one isn't working in the orignal equation either >.<

 

[xeno_gear]

Homework Statement

e^(4x-5)=9e^(x+5)

Homework Equations

log rules

The Attempt at a Solution

I have tried this one a few times using slightly different methods and getting 2 answers and neither of them seem to be working when i plug them back into the equation.

Here is my first method...

1. Divide out the e^(x+5) so i get e^(4x-5)/e^(x+5) = 9

2. Take the natural log of everything so i end up with

(4x-5)lne/(x+5)lne = ln 9

This part is incorrect. The natural log doesn't quite work like that. Before you take natural logs, you'll want to simplify:
$$\frac{e^{4x-5}}{e^{x+5}} = e^{(4x - 5) - (x + 5)} = e^{3x - 10}$$
Then you have e^(3x-10) = 9, and you can take the ln of both sides there.

3. After doing the algebra i got the answer to be 8.87 (rounded to 3 sig figs), but it wasnt checking out when i plugged it back into the equation

Here is my second method...

1. I just started going crazy with the natural logs getting

(4x-5)lne - (x+5) ln 9e

The ln9e don't quite sit right with me tho, however after doing the algebra i get x=-10, this doesn't seem to check out either...please help!

Homework Statement

(3^7x)(27^x)=9

Homework Equations

logs

The Attempt at a Solution

Alright well since i know that they are all powers of 3 i changed the bases so the equation became

(3^7x)((3^3)x)=3^2

Then i just looked at the exponents and ended up with

21x^2 = 2

Seems a little fishy here. You changed everything to base 3 right, but then you should get:
$$3^{7x} \cdot (3^3)^x = 3^{7x} \cdot 3^{3x} = 3^{7x + 3x} = 3^{10x}$$

So then 3^(10x) = 3^2, which means x = ...

Did the algebra and this one isn't working in the orignal equation either >.<

 

[xortan]

Thank you so much it just clicked..

I LOVE this site, got my finals right around the corner and this was only problem i was having with exponentioals, thank you i should be able to complete the rest of my assignment with ease :D

 

[xeno_gear]

aye, no problem.. good luck!

 

[Mentallic]

Just be sure to remove that "o" from "exponentioal" if you want full marks