What is the answer to solvefor p,2^p-2^q=2016 ?

The answer to solvefor p,2^p-2^q=2016 is p=(ln(2016+2^q))/(ln(2))

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solvefor p,2^p-2^q=2016

solve for

p, 2^{p} - 2^{q} = 2016

p = \frac{ln ( 2016 + 2 ^{q} )}{ln ( 2 )}

Solution steps

2^{p} - 2^{q} = 2016

Move

2^{q}

to the right side

2^{p} = 2016 + 2^{q}

Apply exponent rules

p ln (2) = ln (2016 + 2^{q})

Solve

p ln (2) = ln (2016 + 2^{q}) : p = \frac{ln ( 2016 + 2 ^{q} )}{ln ( 2 )}

p = \frac{ln ( 2016 + 2 ^{q} )}{ln ( 2 )}

8 1^{2 x - 3} = 9^{- x}

2^{x} + 2^{x - 1} + 2^{x + 2} = 176

- 1 \cdot 2^{x - 7} + 9 = 3

4^{- x} = 2^{x + 4}

3^{200} + 3^{200} + 3^{200} = 3^{x}

What is the answer to solvefor p,2^p-2^q=2016 ?
The answer to solvefor p,2^p-2^q=2016 is p=(ln(2016+2^q))/(ln(2))