Thread: 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 =)

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

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

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

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

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