algebra help?

Question

"Solve for each exponential equation. If an answer is not exact, give the answer to four decimal places."

1) 4^x+2 - 4^x = 15

2) 3^x+3 +3^x = 84

the 1st problem is supposed to be "x=1," cause it says so in the back of the book. I tried using logs to get rid of exponentials in x+2 and x, but I got a negative decimal. Help?

Answer 1

1) 4^x+2 - 4^x
= 4^x * 4^2 - 4^x
=16 * 4^x - 4^x
= 15* 4^x = 15 (as given)
so 4^x = 1
so x=0 because any integer ^ 0 =1
Check:
if we substitute x =0 in the given equation
4^2 - 4^0 = 16-1 =15 so OK
2) 3^x+3 + 3^x
= 3^x * 3^3 + 3^x
= 27 * 3^x + 3^x
= 28* 3^x = 84 (as given)
dividing both sides by 28
3^x =3
so x=1 as any integer ^ 1 is the same number
Check
if we substitute x =1 in the given equation
3^4+3^1= 81+3 =84 so OK

Answer 2

Is it supposed to be 4^(x+2)? It can't be what you've written because the two 4^x terms would cancel leaving 2=15.

If it's 4^(x+2), the book is wrong, because 4^3 - 4^1 = 60, not 15.
4^(x+2)
= (4^x)*(4^2)
= 16 *(4^x)
and so the equation becomes
15*(4^x) = 15
4^x = 1
x = 0, and if you sub back in the equation you see this is correct.

2)

Answer 3

I am guessing you mean

4^(x+2) - 4^x = 15
4^x(4^2 - 1) = 15
4^x(15) = 15
4^x = 1
x = log[base4](1)
x = 0

3^x+3 +3^x = 84
3^x(3^3 + 1) = 84
3^x(27 + 1) = 84
3^x(28) = 84
3^x = 3
x = log[base3](3)
x = 1

Answer 4

ın the fırst problem x equals zero
and ın the second equatıon x equals 1
1 solutıon
4^x*4^2-4^x=15
4^x(16-1)=15
4^x=1
so x=0
2 solutıon
3^x*3^3+3^x=84
3^x(27+1)=84
3^x=84/28
3^x=3
so x=1

Answer 5

1)
4^(x + 2) - 4^x = 15
(4^x)(4^2) - 4^x = 15
4^x(16 - 1) = 15
4^x = 1
x = 0

2)
3^(x+3) +3^x = 84
3^x(3^3 + 1) = 84
3^x(27 + 1) = 84
3^x = 3
x = 1

Answer 6

you can solve it in this way: 4^x+2-4^x=15 then 4^2*4^x-4^x=15
then 15*4^x=15 then x=0. and exactly in the same way for the second one you can say 27*3^x=84 then 3^x=3 and the x=1

Answer 7

#1

I am guessing that you mean 4^(x+2) - 4^x = 15.

4^(x+2) - 4^x = 15
(4^x)(4^2) - 4^x = 15
16(4^x) - 4^x = 15
15(4^x) = 15
4^x = 1
x = 0


#2

I am guessing that you mean 3^(x+3) + 3^x = 84.

3^(x+3) + 3^x = 84
(3^x)(3^3) + 3^x = 84
27(3^x) + 3^x = 84
28(3^x) = 84
3^x = 3
x = 1