equation

[rattan123]

2^a + 3^b = 17
2^a+2 -3^b+1 = 5
Find the value of a and b.

[imskpwr]

2^a+2 -3^b+1 = 5......2
4*2^a - 3*3^b = 5.....(from 2).........................3
2^a + 3^b = 17.........................................1
(1)*3 and equating with (3), we get

7 * 2^a = 56
=> 2^a = 8 = 2^3
=> a = 3

3^b = 9 = 3^2
=> b=2

Hope this helps!

[rattan123]

Can You explain that how did you equate it..!!

[imskpwr]

Can You explain that how did you equate it..!!

let,
X = 2^a
Y = 3^b

SO equations 1 AND 3 will look like linear.

4X - 3Y = 5..................3
X + Y = 17...................1

HOPE THIS IS CLEAR NOW.

[Anurag@Gurome]

2^(a+2) = (2^a)*(2^2) = 4*(2^a)
3^(b+1) = (3^b)*(3^1) = 3*(3^b)

Hence, 2^(a+2) - 3^(b+1) = 4*(2^a) - 3*(3^b) = 5 ............................. (A)
Multiplying the other equation by 3, 3*(2^a) + 3*(3^b) = 3*17 = 51 ........... (B)
Adding (A) and (B), 7*(2^a) = (51 + 5) = 56 ---> (2^a) = 8 = 2^3 ---> a = 3
Now, (3^b) = 17 - (2^3) = 17 - 8 = 9 = 3^2 ---> b = 2

[rattan123]

Sorry a
change in the question
2^a + 3^b = 17
2^a+b -3^b+1 = 5
Find the value of a and b.

The answer is correct..!!

[Anurag@Gurome]

change in the question
2^a + 3^b = 17
2^a+b -3^b+1 = 5
Find the value of a and b.

From the first equation, (3^b) = 17 - (2^a)
Now, 3^(b+1) = (3^b)*(3^1) = 3*(3^b) = 3*(17 - (2^a)) = 51 - 3*(2^a)
And, 2^(a+b) = (2^a)*(2^b)

--> 2^(a+b) - 3^(b+1) = 5
--> (2^b)*(2^a) - [51 - 3*(2^a)] = 5
--> (2^b)*(2^a) + 3*(2^a) = 56
--> (2^a)[(2^b) + 3] = 56

Now unless it is mentioned a and b are positive integers, it is beyond the scope of GMAT syllabus to find the value of a and b from this.

If we assume a and b are positive integers, then both (2^a) and (3^b) will be integer. Now, 56 can expressed as a product of two integers in following ways : 1*56, 2*28, 4*14, 7*8

Only 7*8 can be written in the form (2^a)[(2^b) + 3] for a = 3 and b = 2.

[GMATGuruNY]

2^a + 3^b = 17
2^a+2 -3^b+1 = 5
Find the value of a and b.

List powers of 2 and 3 and look for a combination with a sum of 17.
2^a = 2^0=1, 2^1=2, 2²=4, 2³=8, 2^4=16.
3^b = 3^0=1, 3^1=3, 3²=9.

Red combination: a=3 and b=2.
Plugging these values into 2^(a+2) - 3^(b+1) = 5, we get:
2^(3+2) - 3^(2+1) = 5.
32-27 = 5.
Success!

Thus, a=3 and b=2.

[confuse mind]

I suppose we will be having proper brackets in the actual exam, otherwise it is difficult to know the actual exponent in this question:

2^a+2 -3^b+1 = 5



No?