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
tan2x*cotx - 3 = 0
We know that: tan2x = sin2x/cos2x and cotx = cosx/sinx
==> sin2x/cos2x *cosx/sinx = 3
Now we know that sin2x = 2sinx*cosx
==> 2sinxcosx/cos2x * cosx/sinx = 3
Reduce sinx:
==> 2cos^2 x/ cos2x = 3
Now we know that cos2x = 2cos^2 x-1
==> 2cos^2 x/(2cos^2 x -1) = 3
==> 2cos^2 x = 3(2cos^2 x -1)
==> 2cos^2 x = 6cos^2 x - 3
==> -4cos^2 x= -3
==> 4cos^2 x = 3
==> cos^2 x = 3/4
==> cosx = +-sqrt3/ 2
<span>==> x = pi/6, 5pi/6, 7pi/6, and 11pi/6</span>
Answer:
Yes
Step-by-step explanation:
y = 5x
The remainder of the polynomial equation is -49
<h3>What is a remainder of a polynomial equation?</h3>
The remainder of a quantity in an algebraic expression of a polynomial equation is the remaining amount of the quantity left and can not be divisible by the divisor.
From the given information, we are to find the remainder of the given algebraic equation:
Using the long division method, we have:
Divide by , we have:
Divide by , we have:
Learn more about finding the remainder of a polynomial equation here:
brainly.com/question/24978997
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Answer:
12
Step-by-step explanation:
f(14)= f (3 + 11)
= f(3) = 12
Therefore the value of f(14) is 12.