# Simple exponential equations?

• Apr 9th 2007, 07:45 AM
erika
Simple exponential equations?
Solve the following equations:

a) 3√(2*4^x) = 8^(1-x)
I worked it out to give 2^(2x+1) = 2^(9-9x) and thus x = 8/11.
Is that correct?

b) 16^x = 2 (2^(2x-1)) + 2

Additionally, if both sides of the equation cannot be expressed as the same base, how do we solve for x?

Step-by-step help would be very much appreciated =)
• Apr 9th 2007, 11:34 AM
topsquark
Quote:

Originally Posted by erika
Solve the following equations:

a) 3√(2*4^x) = 8^(1-x)
I worked it out to give 2^(2x+1) = 2^(9-9x) and thus x = 8/11.
Is that correct?

[2 * 4^x]^{1/3} = 8^{1 - x} <-- Cube both sides

2 * 4^x = [8^{1 - x}]^3

2 * 4^x = 8^{3(1 - x)}

2 * [2^2]^x = [2^3]^{3(1 - x)}

2*2^{2x} = 2^{9(1 - x)}

2^{2x + 1} = 2^{9(1 - x)}

Equating exponents:
2x + 1 = 9(1 - x)

2x + 1 = 9 - 9x

11x = 8

x = 8/11

So yes, you are correct!

-Dan
• Apr 9th 2007, 11:34 AM
Jhevon
Quote:

Originally Posted by erika
Solve the following equations:

a) 3√(2*4^x) = 8^(1-x)
I worked it out to give 2^(2x+1) = 2^(9-9x) and thus x = 8/11.
Is that correct?

Yes, you are correct:)

Quote:

b) 16^x = 2 (2^(2x-1)) + 2

Additionally, if both sides of the equation cannot be expressed as the same base, how do we solve for x?

16^x = 2 (2^(2x-1)) + 2
=> 2^4x = 2^(2x) + 2
=> 2^(2x)2 = 2^(2x) + 2
=> 2^(2x)2 - 2^(2x) - 2 = 0 .......notice that this is a quadratic in 2^(2x)
=> (2^(2x))^2 - (2^(2x)) - 2 = 0 .....if you are the type that will get confused, you can replace 2^(2x) with some variable, say x, and it will look like a regular quadratic.

=> ((2^(2x)) - 2)((2^(2x)) + 1) = 0
=> ((2^(2x)) - 2) = 0 or ((2^(2x)) + 1) = 0
=> 2^(2x) = 2 or 2^(2x) = -1
=> 2^(2x) = 2 since 2^(2x) cannot be -1
equating the powers we get
2x = 1
=> x = 1/2
• Apr 9th 2007, 11:39 AM
topsquark
Quote:

Originally Posted by erika
b) 16^x = 2 (2^(2x-1)) + 2

16^{x} = 2 * (2^{2x - 1}) + 2

[2^4]^{x} = 2^{2x} + 2

2^{4x} = 2^{2x} + 2

2^{4x} - 2^{2x} - 2 = 0

[2^{2x}]^2 - [2^{2x}] - 2 = 0

Define y = 2^{2x}. Then
y^2 - y - 2 = 0

(y - 2)(y + 1) = 0

Thus y = 2 or y = -1.

So
2^{2x} = 2 ==> 2x = 1 ==> x = 1/2
or
2^{2x} = -1, which is impossible.

Thus x = 1/2.

-Dan
• Apr 9th 2007, 11:45 AM
topsquark
Quote:

Originally Posted by erika
Additionally, if both sides of the equation cannot be expressed as the same base, how do we solve for x?

In general, you don't. However there are occasional special cases where you can. For example:
Solve for x:
(2^x)*(3^x) = 216

(2*3)^x = 216

6^x = 6^3

Thus x = 3.

So keep an eye out for ones you can do.

-Dan