Exponents and Logarithms, equation

[sallyj92]

Hey guys I need help how to solve this equation... Express your answer to the equation in the form alnb

9e^4x-e^2x=0

This is as far as I got
9e^4x=e^2x
ln(9e^4x)=ln(e^2x)

the answer given in the markscheme is x=1/2ln1/9, x=-1/2ln9, x=ln1/3, a=-1/2 and b=9, x=-ln3 (accept a=-1 and b=3)

I don't understand why there are so many options? and which ones are right?

 

[Char. Limit]

I recommend letting u=e^x, and then you can express the equation as a quartic. That's why you have so many solutions.

 

[Mark44]

Hey guys I need help how to solve this equation... Express your answer to the equation in the form alnb

9e^4x-e^2x=0

This is as far as I got
9e^4x=e^2x
ln(9e^4x)=ln(e^2x)

the answer given in the markscheme is x=1/2ln1/9, x=-1/2ln9, x=ln1/3, a=-1/2 and b=9, x=-ln3 (accept a=-1 and b=3)

I don't understand why there are so many options? and which ones are right?

Where do a and b come from? They aren't in the original problem.

There are only two solutions. Some of the solutions you give are not distinct. For example, x = 1/2 ln(1/9) is the same as (equal to) ln((1/9)^1/2) = ln(1/3). This is also the same as -ln(3).

The equation can be factored.
9e^(4x) - e^(2x) = 0
<==> e^(2x)(9e^(2x) - 1) = 0

Since e^(2x) is always > 0, the solutions come from 9e^(2x) - 1 = 0, and this in turn can be factored to give the two solutions of the original equation.