Question

Solve the following equations:-
2ᵃ + 3ᵇ = 17
2ᵃ⁺² - 3ᵇ⁺¹ = 5

Answer 1

Hi ,

Let 2^a = x , 3^b = y

i ) 2^a + 3^b = 17

x + y = 17

x = 17 - y --( 1 )

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

=> ( 2^a × 2² ) - ( 3^b × 3¹ ) = 5

[ since , a^m+n = a^m × a^n ]

4x- 3y = 5 ---( 2 )

substitute x = 17 - y in equation ( 2 )

4( 17 - y ) - 3y = 5

68 - 4y - 3y = 5

-7y = 5 - 68

-7y = - 63

y = ( -63 )/( -7 )

y = 9

substitute y = 9 in equation ( 1 ) , we get

x = 17 - 9

x = 8

Therefore ,

2^a = x = 8

=> 2^a = 2³

a = 3 [ since , if a^m = a^n then m = n ]

3^b = y = 9

=> 3^b = 3²

b = 2

Therefore ,

a = 3 , b = 2

I hope this helps you.

: )

Answer 2

HELLO DEAR,

2^a + 3^b = 17

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

[put 2^a = p , 3^b = q]

so, p + q = 17---------( 1 )

4p - 3q = 5-----------( 2 )

from-------------( 1 ) &------------( 2 )

multiply----( 1 ) by "3"

3p + 3q = 51

4p - 3q = 5

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

7p = 56

p = 8 [ put in ----( 1 )]

p + q = 17

8 + q = 17

q = 9

therefore, p = 2^a = 8

2^a = 2^3

a = 3

AND,

q = 3^b = 9

3^b = 3^2

b = 2

hence, a = 3 , b = 2

I HOPE ITS HELP YOU DEAR,

THANKS