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

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: If 2^x = 12 and 6^y = 12, then what is (1/x) + (1/y) equal to?      Log On


   

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

Answer: 1

Explanation

Let's isolate x so we can determine 1/x.
2%5Ex+=+12

log%28%282%5Ex%29%29+=+log%28%2812%29%29 Apply logs to both sides.

x%2Alog%28%282%29%29+=+log%28%2812%29%29 Use the rule log(M^N) = N*log(M) to pull down the exponent.

x+=+log%28%2812%29%29%2Flog%28%282%29%29 Divide both sides by log(2).

1%2Fx+=+log%28%282%29%29%2Flog%28%2812%29%29 Apply the reciprocal to both sides.


6%5Ey+=+12 can be rearranged into 1%2Fy+=+log%28%286%29%29%2Flog%28%2812%29%29 when following similar steps.

Then,


1%2Fx+%2B+1%2Fy+=+%28+log%28%282%29%29++%2B+log%28%286%29%29+%29%2Flog%28%2812%29%29

1%2Fx+%2B+1%2Fy+=+%28+log%28%282%2A6%29%29+%29%2Flog%28%2812%29%29 Use the rule log(M)+log(N) = log(M*N) where the base of each log must be the same.

1%2Fx+%2B+1%2Fy+=+log%28%2812%29%29%2Flog%28%2812%29%29

1%2Fx+%2B+1%2Fy+=+1

-------------------------------------
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

-------------------------------------

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.