Can you find a value for a and b when 2^a = 3^b?

Question

a>0 and b>0
a and b are intergers

Answer 1

No, this cannot be done with integer values only, except for when a = 0 and b = 0. Since a and b are defined to be positive integers (a > 0) and (b > 0), this problem has no solution.

If two positive integers are equal (such as 2^a and 3^b), then their prime factorizations must be equal. 2^a is made up of "a" factors of 2, and 3^b is "b" factors of 3. Since the prime factorizations are not equal, 2^a cannot equal 3^b.

Answer 2

I'll have to work out the details, but I would use the following strategy as a first attempt in solving the problem:


2^a = 3^b implies that a/b = ln(3)/ln(2).

2^a = 3^b, with a, b >0 with both a and b integers, has a solution if, and only if, ln(3)/ln(2) is rational.

Informal Proof:

This is because all rational numbers can be represented as fractions, aka as the ratio of two integers, where as, irrational numbers can not be represented as the ratio of two integers. So, if ln(3)/ln(2) is irrational, this implies that there are no integers a,b > 0 that satisfy the equation
2^a = 3^b.

A side point: there does exist integers a, and b, greater than 0, such that 2^a + 3^b = a perfect square, that is, 2^2 + 3^2 = 5^2. Here a = b = 2. Hope this helps.

Answer 3

A more formal writing of the good solution above..

2 ^ a = 0 ^ a mod 2 = 0 mod 2

3 ^ b = 1 ^ b mod 2 = 1 mod 2

Clearly they can never be the same.

Answer 4

The only values that fit are a=0 and b=0. Anything to the 0 power = 1.

Answer 5

There is no integer solution set.

All powers of 3 end in the digits 3, 9, 7, or 1.

All powers of 2 end in the digits 2, 4, 8, or 6.

So 2 to any integer power can NEVER equal 3 to any integer power. (Except zero, but you have ruled that one out.)

Answer 6

With the restriction of a, b being positive, non-zero integers there is no solution for 2^a = 3^b.

However, if you let a, b be elements of the Reals, then for any b, a= b*ln(3)/ln(2) would be a solution.

Answer 7

a = b log 3 to the base 2
b = a / (log 3 to the base 2)

try any number for a then you could solve for b

and believe me this works...

example:

let b = 3

a = b log 3 to the base 2
a = 3 log 3 to the base 2
a = 3 (1.5849625007211561814537389439478...)
a = 4.7548875021634685443612168318434...

after getting the values all we have to do now is substitute

2^a = 3^b
2^(4.7548875021634685443612168318434...) = 3^3
27 = 27

see it's correct

^_^

Answer 8

No.
At the most we can find a/b from the following:

2^a = 3^b

a log 2 = b log 3
a/b = log 3 / log 2 = 1.5849625

Answer 9

If a and b are both integers, then 2^a = 3^b have no soultions.

2^a is always even. 3^b is always odd.

Answer 10

a = 3^b/log 2
b = 2^a/log 3