SOLUTION: x=log a base 2a, y=log 2a base 3a, z=log 3a base 4a then prove that xyz + 1=2yz

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)

-----

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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