# Exponential equation

• Mar 16th 2014, 01:18 AM
andy000
Exponential equation
Hi,

I am really stuck on how you deal with with an exponential equation something like this:

x^y - 5 * x^y = 0 I know I cant multiply 5 * x^y as they have a different base (right?)

How do i simplify this and get the variables/exponents together?

Thanks
• Mar 16th 2014, 01:43 AM
earboth
Re: Exponential equation
Quote:

Originally Posted by andy000
Hi,

I am really stuck on how you deal with with an exponential equation something like this:

x^y - 5 * x^y = 0 I know I cant multiply 5 * x^y as they have a different base (right?)

How do i simplify this and get the variables/exponents together?

Thanks

Good morning,

is this the correct equation?

$x^y - 5x^y = 0$

If so, collect like terms:

$-4x^y = 0~\implies~x^y = 0$

There is only one real solution : $x = 0~ \wedge~ y \ne 0$
• Mar 16th 2014, 01:54 AM
andy000
Re: Exponential equation
That is not the correct equation.

The actual equation i am working on is:

16^x - 5 * 8^x = 0

I have tried converting into base of 2 so:

(2^4)^x - 5 * (2^3)^x = 0

= 2^4x - 5 * 2^3x = 0

but I have no idea how to deal with the -5 * part of the equation.

Thanks
• Mar 16th 2014, 02:00 AM
earboth
Re: Exponential equation
Quote:

Originally Posted by andy000
That is not the correct equation.

The actual equation i am working on is:

16^x - 5 * 8^x = 0

I have tried converting into base of 2 so:

(2^4)^x - 5 * (2^3)^x = 0

= 2^4x - 5 * 2^3x = 0

but I have no idea how to deal with the -5 * part of the equation.

Thanks

Hello,

you can factor the term on the LHS:

$16^x - 5 \cdot 8^x = 0 ~\implies~2^{3x} \left(2^x-5 \right)=0$

A product is zero if one factor is zero. Since $2^{3x} \ne 0$ only

$2^x-5 = 0~\implies~2^x=5~\implies~x = \log_2(5)$
• Mar 16th 2014, 02:08 AM
andy000
Re: Exponential equation
Hi, thanks very much!

When you write 2^3x does not equal 0 do you mean 2^3x could not equal 0 for any value of x thus we can't get a real solution from that part of the equation?

Thanks again
• Mar 17th 2014, 09:55 AM
bjhopper
Re: Exponential equation
another approach
16^x= 5*8^x
xlog 16 =log 5 + xlog8
using base 2
4x=log 5 +3x
x=log base 2 (5)
x = ln5 /ln2
• Mar 17th 2014, 10:55 AM
JeffM
Re: Exponential equation
Quote:

Originally Posted by andy000
Hi, thanks very much!

When you write 2^3x does not equal 0 do you mean 2^3x could not equal 0 for any value of x thus we can't get a real solution from that part of the equation?

Thanks again

Yes.

$x\ is\ a\ real\ number \implies 2^x > 0 \implies \left(2^x\right)^3 > 0 \implies 2^{3x} > 0 \implies 2^{3x} \ne 0.$

I think the easiest way to deal with this is:

$16^x - 5 * 8^x = 0 \implies 16^x = 5 * 8^x \implies$

$\left(2^4\right)^x = 5 * \left(2^3\right)^x \implies$

$2^x = 5 \implies$

$ln(2^x) = ln(5) \implies$

$x = \dfrac{ln(5)}{ln(2)}.$

Edit: The answer in terms of log base 2 is more succinct, but it does not readily lead to a numeric approximation whereas an answer in terms of log base e can easily be computed on a scientific calculator.