PLEASE HELP ME on #57 WITH ALL WORK SHOWN!!

Tmr is my exam pls help...

Explanation:

Recall that quadratic form is ax^2+bx+c.

So we need an x^2 term followed by an x term, then a constant at the end.

It may not look it, but the given equation is nearly in quadratic form.

Rewrite 9^(2x) as (9^x)^2

We get this equivalent equation (9^x)^2 - 2*(9^x) - 3 = 0

From here, let's make w = 9^x. Replace every copy of '9^x' with 'w' and we end up with this new simpler equation: w^2 - 2w - 3 = 0

From here we can use the quadratic formula or factor. I'll factor.

w^2 - 2w - 3 = 0

(w - 3)(w + 1) = 0

w-3 = 0 or w+1 = 0

w = 3 or w = -1

Now we use these values of w to find x

If w = 3, then,

w = 9^x

3 = 9^x

log(3) = log(9^x)

log(3) = x*log(9)

x = log(3)/log(9)

x = log(3)/log(3^2)

x = log(3)/(2*log(3))

x = 1/2

At the last step, we have the log(3) terms cancel.

Repeat for w = -1

w = 9^x

-1 = 9^x

log(-1) = log(9^x)

We run into a problem. The log of any negative number is not a real number output. In other words, 9^x is never negative. Raising any positive number to a real number output leads the result to be positive. Take a look at the graph of y = 9^x to confirm this.

In short, w = -1 is extraneous so it doesn't lead to any x value solution.

So only w = 3 is useful.

Ultimately, the only solution is x = 1/2.

answer: 20

step-by-step explanation:

step-by-step explanation: