Question

How do you solve `3^(2x) – 2*3^(x+5) + 3^10 = 0`?

Answer 1

`x=5`

Explanation:

Remember the polynomial identity
`(a+b)^2=a^2+2ax+b^2`. Choosing `a=3^x` and `b=-3^5` we have
`(3^x-3^5)^2=3^{2x}-2*3^x*3^5+3^{10}`. So, our problem is reduced to: Solve for `x` the condition `(3^x-3^5)^2=0->3^x-3^5=0->3^x=3^5->x=5`

Answer 2

x=5

Explanation:

At first sight this seems a quadratic equation of the form:
`y^2+by+c` where `y=3^x`. So let's transform the equation into the quadratic form:

`3^(2x)-2xx3^5xx3^x+3^10`

We can see that the equation can also be reduced to

`(3^x)^2-2xx3^5xx3^x+(3^5)^2`

Remember that :

`color(blue)(a^2-2ab+b^2=(a-b)^2)`

The equation can be arranged into:

`(3^x-3^5)^2=0`

So:

`3^x-3^5=0`

`3^x=3^5`

`x=5`