What is the answer to solvefor x,4^x+5^y=21 ?

The answer to solvefor x,4^x+5^y=21 is x=(ln(21-5^y))/(2ln(2))

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solvefor x,4^x+5^y=21

solve for

x, 4^{x} + 5^{y} = 21

x = \frac{ln ( 21 - 5 ^{y} )}{2 ln ( 2 )}

Solution steps

4^{x} + 5^{y} = 21

Move

5^{y}

to the right side

4^{x} = 21 - 5^{y}

Apply exponent rules

x ln (4) = ln (21 - 5^{y})

Solve

x ln (4) = ln (21 - 5^{y}) : x = \frac{ln ( 21 - 5 ^{y} )}{2 ln ( 2 )}

x = \frac{ln ( 21 - 5 ^{y} )}{2 ln ( 2 )}

m, x^{m - 1} = 0

7^{x - 3} \cdot 98 = 14

e^{y} = 11

e^{y} = 12

8^{1 + x} + 8^{1 - x} = 20

What is the answer to solvefor x,4^x+5^y=21 ?
The answer to solvefor x,4^x+5^y=21 is x=(ln(21-5^y))/(2ln(2))