(x^a)(x^b)=x^(a+b)
(ab)(cd)=(a)(b)(c)(d)
x^-m=1/(x^m)
(3y^-4)(2y^-4)=
(3)(y^-4)(2)(y^-4)=
(6)(y^-8)=
6/(y^8)
You can take the log of the left and right hand side, and then apply the <span>logarithm rules:
log(a</span>ˣ) = x·log(a)
log(ab) = log(a) + log(b)
log(9^(x-1) * 2^(2x+2)) = log(6^(3x))
log(9^(x-1)) + log(2^(2x+2)) = 3x log(6)
(x-1) log(9) + (2x+2) log(2) - 3x log(6) = 0
x(log9 + 2log2 - 3log6) = log9 - 2log2
x = (log9 - 2log2) / (log9 + 2log2 - 3log6)
simplifying by writing log9 = 2log3 and log6 = log2+log3
x= 2(log3 - log2) / (2log3 + 2log2 - 3log2 - 3log3) =
x= -2(log3 - log2) / (log3 + log2) = -2 log(3/2) / log(6)
So 6^x = 4/9
Answer:
see explanation
Step-by-step explanation:
the equation of parabola in vertex form is
y = a(x - h)² + k
where (h, k ) are the coordinates of the vertex and a is a multiplier.
here (h, k ) = (3, 1 ) , then
y = a(x - 3)² + 1
to find a substitute any other point on the graph into the equation.
using (0, 7 )
7 = a(0 - 3)² + 1 ( subtract 1 from both sides )
6 = a(- 3)² = 9a ( divide both sides by 9 )
=
= a
y =
(x - 3)² + 1 ← in vertex form
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the equation of a parabola in factored form is
y = a(x - a)(x - b)
where a, b are the zeros and a is a multiplier
here zeros are - 1 and 3 , the factors are
(x - (- 1) ) and (x - 3), that is (x + 1) and (x - 3)
y = a(x + 1)(x - 3)
to find a substitute any other point that lies on the graph into the equation.
using (0, - 3 )
- 3 = a(0 + 1)(0 - 3) = a(1)(- 3) = - 3a ( divide both sides by - 3 )
1 = a
y = (x + 1)(x - 3) ← in factored form
Answer:
What's the question?
Step-by-step explanation: