Solve the equations
2^x – 5^y = 3 and 2^x-3 + 5 ^y-2 =21​

You can substitute a = 2ˣ and b = 5ʸ to make both equations linear.

The first equation becomes

a - b = 3

In the second equation, first simplify the exponents:

2ˣ ⁻³ = 2ˣ × 2⁻³ = 1/8 × 2ˣ

5ʸ ⁻² = 5ʸ × 5⁻² = 1/25 × 5ʸ

so the second equation changes to

a/8 + b/25 = 21

Multiply both sides of this by 200 (the LCM of 8 and 25) to eliminate the fractions:

25a + 8b = 4200

Multiply both sides of the first equation by 8, then add the corresponding sides of both equations. This eliminates b and you can solve for a :

8 (a - b) + (25a + 8b) = 8 (3) + 4200

(8a - 8b) + (25a + 8b) = 24 + 4200

33a = 4224

a = 128

Plug this into the first equation to solve for b :

128 - b = 3

b = 125

Now solve for x and y. Notice that

128 = 2⁷

and

125 = 5³

so it follows that x = 7 and y = 3.

Alternatively, you can use logarithms. For example,

a = 2ˣ   →   log₂(a) = log₂(2ˣ )   →   x = log₂(128) = 7

none of these

step-by-step explanation:

23 + 34 add up to 57 so just inform the teacher of this really stupid mistake

b.

step-by-step explanation: