3^a = 9^b = 27^c and a, b, and c don’t equal 0, what is (a)/(b) + (b)/(c) + (c)/(a)

Question

`3^a = 9^b = 27^c` and a, b, and c don’t equal 0, what is
`(a)/(b) + (b)/(c) + (c)/(a)`

Answer 1

To solve the expression
`\large\sf\:(a)/(b) + (b)/(c) + (c)/(a)\\` given the conditions
`\large\sf\:3^a = 9^b = 27^c\\`, we can use logarithmic properties and the fact that
`\large\sf\:3^2 = 9\\` and
`\large\sf\:3^3 = 27\\`.

Let's start by finding the values of a, b, and c using logarithmic properties:

Taking the logarithm of both sides of
`\large\sf\:3^a = 9^b\\`, we get:

`\large\sf\:\log_3(3^a) = \log_3(9^b)\\`

Applying the power rule of logarithms, we can bring down the exponents:

`\large\sf\:a\log_3(3) = b\log_3(9)\\`

Since
`\large\sf\:\log_3(3) = 1` and
`\large\log_3(9) = 2\\`, we simplify to:

`\large\sf\:a = 2b\\` ---- (1)

Similarly, taking the logarithm of both sides of
`\sf\:9^b = 27^c\\`, we get:

`\large\sf\:b\log_3(9) = c\log_3(27)\\`

Using the values of
`\sf\:\log_3(9)\\` and
`\sf\:\log_3(27)\\` as before, we have:

`\large\sf\:b(2) = c(3)\\`

Simplifying, we get:

`\large\sf\:2b = 3c\\` ---- (2)

Now, let's substitute the value of b from equation (1) into equation (2):

`\large\sf\:2(2b) = 3c\\`

`\large\sf\:4b = 3c\\`

Rearranging, we find:

`\large\sf\:c = (4b)/(3)\\` ---- (3)

We now have expressions for a, b, and c in terms of b. Let's substitute these into the expression
`\large\sf\:(a)/(b) + (b)/(c) + (c)/(a)\\`:

`\large\sf\:(a)/(b) + (b)/(c) + (c)/(a) = (2b)/(b) + (b)/((4b)/(3)) + ((4b)/(3))/(2b)\\`

Simplifying further, we get:

`\large\sf\:(2)/(1) + (3)/(4) + (2)/(3)\\`

Finding the common denominator and combining the fractions, we have:

`\large\sf\:(24)/(12) + (9)/(12) + (8)/(12)\\`

Adding the fractions together, we obtain:

`\large\sf\:(24 + 9 + 8)/(12) = (41)/(12)\\`

Therefore,
`\large\sf\:(a)/(b) + (b)/(c) + (c)/(a) = (41)/(12)\\`.