Solving Exponential Equations SO CONFUSED ?

[Monocerotis]

Solving Exponential Equations SO CONFUSED ?

Homework Statement

I'm trying to solve problems like these
a) 2(3^y-2) = 18
b) 27(3^3x+1) = 3
c) 2^x+2 - 2^x = 48
d) 10^z+4 + 10^z+3 = 11

Homework Equations

Laws of exponential values.

The Attempt at a Solution

b) 2(3^y-2) = 18
6^y-2 = 18
Stuck/Confused.

Maybe there's something fundamental I'm forgetting or making overly complicated. If you divide 18 by 3 to make it 6 so then you could just work with the exponential values to solve for y, wouldn't you also have to do the same thing to 2(3^y-2) (or) 6^y-2

For say a problem like this
3(5^x+1) =15
15^x+1 = 15

Now you have both bases balanced, you can work to solve for x
x+1 = 0 ?
x = 0 - 1
x = -1

That would then mean 15⁰ = 15, which doesn't work out because anything to an exponential value of 0 is 1, correct ?

 

[maladroit]

use natural logs to solve for the exponents!
ie 2(3^y-2) = 18
3^y-2=9
ln3^y=ln9
yln3=ln9
y=ln9/ln3

all of them can be solved that way!

 

[Monocerotis]

Ok that sounds interesting. We haven't learned how to use natural logs in class yet, they begin teaching you how to use that in the next level course.

Is there any other way to solve the problem without using natural logs ?

If not I guess I'll just have to teach myself lol

 

[maladroit]

correction!

2(3^y-2) = 18
3^y-2=9
ln3^y-2=ln9
(y-2)ln3=ln9
y=(ln9)+2/ln3

 

[maladroit]

Okay! well, if the exponents have the same base (ie 2^x-1=2^6-x) then you can set the exponents equal to each other. For your problems, you will have to manipulate the bases to get them to be the same... for example if you have 2^x-4=8, that also equals 2^x-4=2^3, so then x-4=3.

 

[maladroit]

Also-Because there are exponents involved, don't multiply the number outside of the parentheses inside of the parentheses!

 

[darkchild]

Ok that sounds interesting. We haven't learned how to use natural logs in class yet, they begin teaching you how to use that in the next level course.

Is there any other way to solve the problem without using natural logs ?
l

Using natural logs is quite easy (it's just another logarithm, with base e), but you can also try re-writing the side of the equation that does not have exponents:

2(3^y-2) = 18

3^y-2 = 9

3^y-2 = 3²

Once both sides have the same base (3 in this case), you can set the exponents equal to solve for the variable:

y-2 = 2

The only problem is that the numbers don't always work out so easily (for example, if you had = 5 instead of = 18, you would need a calculator and end up with horrible decimal numbers).

 

[maladroit]

Excellent explanation!

 

[Bohrok]

If you need to take the log of both sides, you can do it in any base (not just ln or log_e), such as log₁₀ or log₃

 

[Monocerotis]

Thanks Bohrok, maladroit, and darkchild for your help.

Don't know what it is but sometimes I just get stuck on problems and can't work my way out of them.

I think i'll be able to figure it all out now.

 

[mrkuo]

All of these problems can be done without the use of logarithms.

You'll need to be able to factor correctly in the last two to solve that without a calculator.

 

[Monocerotis]

For this example problem that I listed, I know off the bat the answer need to be x = 0
so why when I work it out do I end up with x = -1. What am I doing wrong in the problem solving process ?

3(5^x+1) =15
15^x+1 = 15
--------------------
^x+1 = 0 ?
x = 0 - 1
x = -1

 

[Mark44]

For this example problem that I listed, I know off the bat the answer need to be x = 0
so why when I work it out do I end up with x = -1. What am I doing wrong in the problem solving process ?

3(5x+1) =15
15x+1 = 15

3(5x + 1) != 15x + 1.
Hint: Distributive law.

--------------------
x+1 = 0 ?
x = 0 - 1
x = -1

 

[mrkuo]

If this is the problem you're referring to:

3(5^(x+1)) =15
15^(x+1) = 15

You can not distribute 3 to 5 since they have different exponents. (1 versus x+1)

Instead of distributing, try dividing 3 to both sides. The 15 will become a 5 and you can set the exponents on both sides equal.

 

[Monocerotis]

3(5x + 1) != 15x + 1.
Hint: Distributive law.

I think you may be working of my typo, the x+1 should have been ^x+1 (exponential values). So when 15 ^x+1 = 15 you "ignore" the values of 15 and work to solve for x with the exponential values.

So is my problem then neglecting that 15 = 15¹, so then I would have to work with ^x+1 = ¹
-> x = 1-1 = 0

 

[maladroit]

Don't distribute! Divide both sides by 3, leaving you with 5^x+1=5
and remember that 5=5^1

 

[Monocerotis]

Don't distribute! Divide both sides by 3, leaving you with 5^x+1=5
and remember that 5=5^1

thought so

/facepalm