If $5^{x}-3^{y}=13438$ and $5^{x-1}+3^{y+1}=9686$, then $x+y$ equals

Given:

$5^x - 3^y = 13438$ ... (1)

$5^{x-1} + 3^{y+1} = 9686$ ... (2)

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Key insight: We can rewrite these equations using the fundamental exponent rules:

$5^x = 5^{x-1} \times 5^1 = 5 \times 5^{x-1}$

$3^{y+1} = 3^y \times 3^1 = 3 \times 3^y$

This creates a connection between our two equations that we can exploit.

From equation (1): $5^x - 3^y = 13438$

Substituting $5^x = 5 \times 5^{x-1}$:

$5 \times 5^{x-1} - 3^y = 13438$ ... (3)

From equation (2): $5^{x-1} + 3^{y+1} = 9686$

Substituting $3^{y+1} = 3 \times 3^y$:

$5^{x-1} + 3 \times 3^y = 9686$ ... (4)

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Let's make this simpler by setting:

$a = 5^{x-1}$

$b = 3^y$

This substitution turns our exponential equations into linear equations - much easier to solve!

Our equations become:

From (3): $5a - b = 13438$ ... (5)

From (4): $a + 3b = 9686$ ... (6)

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From equation (5): $5a - b = 13438$

From equation (6): $a + 3b = 9686$

To eliminate $a$, we multiply equation (6) by 5:

$5a + 15b = 48430$ ... (7)

We subtract equation (5) from equation (7):

$(5a + 15b) - (5a - b) = 48430 - 13438$

$5a + 15b - 5a + b = 34992$

$16b = 34992$

$b = 2187$

We substitute $b = 2187$ into equation (6):

$a + 3(2187) = 9686$

$a + 6561 = 9686$

$a = 3125$

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Remember: $a = 5^{x-1}$ and $b = 3^y$

Find x:

$5^{x-1} = 3125$

We recognize powers: $3125 = 5^5$ (since $5^5 = 5 \times 5 \times 5 \times 5 \times 5 = 3125$)

Therefore: $5^{x-1} = 5^5$

This gives us: $x - 1 = 5$

So: $x = 6$

Find y:

$3^y = 2187$

We recognize powers: $2187 = 3^7$ (since $3^7 = 3^6 \times 3 = 729 \times 3 = 2187$)

Therefore: $3^y = 3^7$

This gives us: $y = 7$

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$x + y = 6 + 7 = 13$

Answer: 13