Solve 9x = 27
1/3
2/3
3/2
Use the rules of exponents to evaluate and simplify the following expression. Type all without negative exponents.
(ab)^2 =
If (2, y) lies on the graph of y = 3^x, then y =
9
6
1/9
solve 2^x = 1/32
5
1/5
5
(32)^3/5=
1/8
8
no real number
Solve 2^x = 64
6
7
8
3 Answers

Question 1:
Solve 9x=27
9x = 27
x =27/9
x = 3
Question 2:
(ab)^2
=(1/ab)^2
=1/a^(2)b^(2)
Question 3:
If (2,y) lies on the graph of y = 3x, then y =?
y = 3x
y = 3^(2)
y =1/3^2
y =1/9
Question 4: Solve 2^x =1/32
2^x=1/32
(since 1/32 = 1/2^5 = 2^(5))
2^x=2^5
(if 2^x = 2^y, then x=y)
x=(5)
Question 5:
(32)^(3/5)
=fifth root of [(32)^3]
=fifth root of [32768]
=8
or
(32)^3/5 (no parenthesis)
=(32768)/5
=6553.6 (it’s real, but has a decimal)
Question 6:
2^x = 64
2^x = 64
2^x = 2^6
x = 6
remember that:
(a) to make a power positive, get it’s reciprocal
Ex: (2/5)^(3)
=(5/2)^(3)
=(5^3/2^3)
=125/8
(b) in an exponential fraction, the numerator becomes the exponent of base while the denominator will be its root
Ex: 2^(2/3)
=cube root of 2^2
=cube root of 4
Ex: (3/4)^(2/3)
=(4/3)^(2/3)
=[4^(2/3)/3^(2/3)]
=[cube root of 4^2/cube root of 3^2]
=[cbrt {16}/cbrt {9}]
=cube root of (16/9)
(c) if both sides of an equation has only equal bases with only division or multiplication, you can equate its exponents. remember that in multiplication, exponents are added and in div., exp. are subtracted.
Ex: 3^x=3^5
x=5
Ex: 3^x={(3^5)(3^2)}/3^4
x={5+2}4
x=3
Ex: 3^x=(3^5)+(3^2)
In this case, you cannot equate the exponents automatically, here, you can use logarithm to equate x (its another big talk so don’t mind it first). 🙂
just follow these rules and i’m sure you can perfect your assignments 🙂 🙂 🙂 🙂

9x=27
x=3
if(2,y)
y=3^x
answer 1/9
2^x = 1/32
x=5
(32)^3/5
no real number
2^x = 64
x=8

i basically attempt to unravel this concern : first x=3 the consequence is1 6 10 6 9 0 2nd x=(one million) the consequence is one million 10 18 fifty 4 27 0 the 0.33 x=3 the consequence is one million 3 one million 3 0 the forth x=(3) the consequence is one million 0 one million 0 the final x=(one million) the consequence is one million one million 0 so the roots are (x=3)(x=3)(x=3)(x=one million)(x=one million)