SOLUTION: 5. Solve the following equations
a) 3^m=81
b) (36p^4)12=24
c) 5^n×5^n+4=25
d) 4^7x/256=64^2x+2/16^2x−1
Algebra ->
Square-cubic-other-roots
-> SOLUTION: 5. Solve the following equations
a) 3^m=81
b) (36p^4)12=24
c) 5^n×5^n+4=25
d) 4^7x/256=64^2x+2/16^2x−1
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Question 1005734: 5. Solve the following equations
a) 3^m=81
b) (36p^4)12=24
c) 5^n×5^n+4=25
d) 4^7x/256=64^2x+2/16^2x−1 Found 2 solutions by stanbon, Boreal:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Solve the following equations
a) 3^m=81
3^m = 3^4
m = 4
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b) (36p^4)12=24
36p^4 = 2
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p^4 = 1/18
p = (1/18)*18^(1/4)
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c) 5^n×5^(n+4)=25
5^(2n+4) = 5^2
2n+4 = 2
2n = -2
n = -1
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d) 4^7x/256=64^2x+2/16^2x−1
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(2^2)^(7x)/2^8 = (2^6)^(2x)/(2^4)^(2x-1)
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[2^(14x-8) = [2^(12x)/2^(8x-4)]
======
2^(14x-8) = 2^(4x+4)
-------
14x-8 = 4x+4
10x = 12
x = 1.2
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Cheers,
Stan H.
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You can put this solution on YOUR website! 3^m=81
m=4. One can take logs, but 3^2=9 and 9^2=81.
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(36p^4)12=24
36p^4=2
18p^4=1
18p^4-1=0
This can be p^4=(1/18)
p= +/- (1/18)^(1/4). That is two roots. You get the other complex ones by setting it up this way:
(sqrt(18)p^2+1)(sqrt(18) p^2-1)=0
the second is p^2=(1/sqrt(18) p=+/ - 1/18^(1/4)
The first is p= i +/- 1/18^(1/4)
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5^n×5^n+4=25
5^(2n+4)=25=5^2
Therefore, 2n+4=2, since the bases are equal. If we took logs to base 5, the base would disappear.
2n+4=2
n= -1
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4^7x/256=64^2x+2/16^2x−1
4^7x/4^4=4^(7x-4). That is the left side.
4^3(2x+2)/4^2(2x-1). That is the right side.
4^(7x-4)=4^(6x+6)/4^(4x-2)
log to base 4
7x-4= log 4 4^(6x+6-4x+2)=2x+8
5x=12
x=12/5
This checks.
The left side will be 50859008.46, which is 4^(84/5)-(20/5) or 4^(64/5)
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