If log y = 1.5x-2, show that y = 0.01(31.6)^x
The farthest I got was that 10^(1.5x-2)=y
but then I am stuck
2006-12-20 11:03:11 · 5 answers · asked by alikat4392 4 in Science & Mathematics ➔ Mathematics
Laws of logarithms are not til the next chapter, is there a way to do it without using that log a + log b= log(ab))??
2006-12-20 11:29:27 · update #1
Here's how:
10^(1.5x - 2)
= 10^(1.5x + -2)
= 10^(1.5x) times 10^-2 [because when you multiply you add the exponents]
= (10^1.5 )^x times 0.01 [because when a power is raised to a power you multiply the exponents]
= 0.01(31.6)^x [because 10^1.5 is about 31.6]
= 0.01(10^1.5)^x
2006-12-20 12:04:05 · answer #1 · answered by hayharbr 7 · 0⤊ 0⤋
log(y) = 1.5x - 2
y = 10^(1.5x - 2)
y = 10^(1.5x)/10^2
y = (10^((3/2)x)))/100
(10^((3/2)x)))/100 = .01(31.6)^x
10^((3/2)x) = 31.6^x
(10^(3/2))^x = 31.6^x
31.6^x = 31.6^x
For other logarithm help, go to people.hofstra.edu/faculty/Stefan_Waner/RealWorld/calctopic1/logs.html
2006-12-20 19:50:47 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
log(base 10) y = 1.5x - 2
Exponentiate each side.
y = 10^(1.5x - 2) = [(10^1.5)^x]*10^(-2)
But 10^(1.5) = 31.6 to one decimal place. So
y = (31.6^x)(0.01) = 0.01(31.6)^x
2006-12-20 20:08:12 · answer #3 · answered by Northstar 7 · 0⤊ 0⤋
logy= 1.5x - 2
logy + 2 = 1.5x
logy + log100 = 1.5x (because log100 = 2)
log(100y) = 1.5x (because loga + logb= log(ab))
100y=10^(1.5x)
100y=[10^(1.5)]^x
100y= 31.6^x (because 10^(1.5)=31.6)
y=(1/100)31.6^x
y= 0.01(31.6^x) (because 1/100 = 0.01)
Solved
2006-12-20 19:24:12 · answer #4 · answered by Vu 2 · 2⤊ 0⤋
logy=x
let's take exp. of both sides:
e^logy=e^x or e^logy=y y=e^x
logy=1.5x-2
y=e^(1.5x-2)
set, logy(base10)=1.5x-2 or y=10^(1.5x-2)
y=10^1.5x/10^2
sorry.....
2006-12-20 20:28:29 · answer #5 · answered by Johnny 2 · 0⤊ 0⤋