Question 970605: x=log a base 2a, y=log 2a base 3a, z=log 3a base 4a then prove that xyz + 1=2yz
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you are given:
x = L2a(a)
y = L3a(2a)
z = L4a(3a)
you are asked to prove that xyz+1 = 2yz
note that Lb(x) means log of x to the base of b.
for example: L2a(a) means log(a) to the base of 2a.
you can convert everything to base of 10 using the log base conversion formula of:
Lb(x) = L(x)/L(b)
that means log of x to the base of b is equivalent to log of x to the base of 10 divided by log of b to the base of 10.
your original equations get converted as follows:
x = L2a(a) becomes x = L(a)/L(2a)
y = L3a(2a) becomes y = L(2a)/L(3a)
z = L4a(3a) becomes z = L(3a)/L(4a)
applying these equivalencies, you get:
2yz = 2*L(2a)/L(3a)*L(3a)/L4a) which becomes:
2yz = 2*L(2a)/L(4a).
since 2*L(2a) becomes L((2a)^2) which becomes L(4a^2), then:
2yz = 2*L(2a)/L(4a) bcomes:
2yz = L(4a^2)/L(4a)
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xyz+1 = L(a)/L(2a)*L(2a)/L(3a)*L(3a)/L(4a)+1 which becomes:
xyz+1 = L(a)/L(4a)+1 which becomes:
xyz+1 = L(a)/L(4a)+L(4a)/L(4a) which becomes:
xyz+1 = (L(a)+L(4a))/L(4a)
since L(a)+L(4a) is equal to L(a*4a) which is equal to L(4a^2), then:
xyz+1 = (L(a)+L(4a))/L(4a) becomes:
xyz+1 = L(4a^2)/L(4a).
since 2yz = L(4a^2)/L(4a) and xyz+1 = L(4a^2)/L(4a), then:
2yz = xyz+1
end of proof
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