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how would you calculate these problems i was absent for the lesson and i have no notes! like its 2 squared but instead where the squared would go it would be like

2^2x+2= 16
100^x+1= 1/100
4^6x= 1/64
4^x = 8 cubed
3^9x = 1/81


please help me out greatly appreciated =)

2007-03-12 12:33:46 · 1 answers · asked by Anton d 2 in Education & Reference Homework Help

1 answers

I imagine you are supposed to express both sides as powers of the same base, then set the exponents equal to each other.

First prob: 16 = 2^4
so 2^(2x+2) = 2^4
Set powers = to each other:
2x+2 = 4
2x=2
x=1

2nd prob: 100 = 10^2, and 1/100 = 1/10^2 = 10^(-2)
So 10^2^(x+1) = 10^(-2)
Use the power-to-a-power rule, which says to multiply the powers, so you get
10^(2x+2)= 10^(-2)
Then 2x+2 = -2, so 2x= -4, so x=-2.


3rd prob: 1/64 = 1/4^3 = 4^(-3)
4^6x = 4^(-3)
6x= -3, so x=-1/2

4th: 4 = 2^2, and 8 = 2^3

(2^2)^x = (2^3)^3
2^2x = 2^9
2x=9, so x=9/2

Last one:
81 = 3^4, so 1/81 = 3^(-4)
3^9x = 3^(-4)
9x=-4
x=-4/9

2007-03-12 12:38:45 · answer #1 · answered by jenh42002 7 · 0 0

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