SOLUTION: If 2^x = 12 and 6^y = 12, then what is (1/x) + (1/y) equal to?

Question 1209181: If 2^x = 12 and 6^y = 12, then what is (1/x) + (1/y) equal to?
Answer by math_tutor2020(3816) (Show Source): You can put this solution on YOUR website!

Answer: 1

Explanation

Let's isolate x so we can determine 1/x.

Apply logs to both sides.

Use the rule log(M^N) = N*log(M) to pull down the exponent.

Divide both sides by log(2).

Apply the reciprocal to both sides.

can be rearranged into when following similar steps.

Then,

Use the rule log(M)+log(N) = log(M*N) where the base of each log must be the same.

-------------------------------------
Another approach.

2^x = 12 rearranges to 2 = 12^(1/x) when raising both sides to the 1/x power.
Similarly we can say that 6^y = 12 turns into 6 = 12^(1/y)

We have these new equations
2 = 12^(1/x)
6 = 12^(1/y)

These two new equations involve exponents 1/x and 1/y.
If we multiply straight down then we'll be able to add these exponents due to the rule a^b*a^c = a^(b+c).

The left hand sides multiply to 12 aka 12^1.
The right hand sides multiply to 12^( (1/x) + (1/y) )
We form the equation 12^1 = 12^( (1/x) + (1/y) )

Comparing the two sides shows that the bases are both 12, so the exponents must be equal.

Therefore (1/x) + (1/y) = 1

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Let's confirm the answer.

Use a calculator and the change of base formula to determine these two approximations
x = log₂(12) = log(12)/log(2) = 3.584963
y = log₆(12) = log(12)/log(6) = 1.386853
Each approximate value is rounded to 6 decimal places.

Then,
(1/x) + (1/y) = (1/3.584963) + (1/1.386853)
(1/x) + (1/y) = 0.999999860928
We don't land on 1 exactly but we get close enough.
The rounding error will depend how you rounded the approximations for x and y.

If you were to use more decimal digits in the values of x and y, then you'll get closer to 1.
Let's say we rounded to 12 decimal places (rather than 6)
(1/x) + (1/y) = (1/3.584962500721) + (1/1.386852807235)
(1/x) + (1/y) = 0.9999999999997738
We get much closer to 1 this time.
There are 12 copies of "9" listed in that number shown above.

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