What is the answer to solvefor p,2^{p-1}+5=3^q ?

The answer to solvefor p,2^{p-1}+5=3^q is p=(ln(3^q-5))/(ln(2))+1

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solvefor p,2^{p-1}+5=3^q

solve for

p, 2^{p - 1} + 5 = 3^{q}

p = \frac{ln ( 3 ^{q} - 5 )}{ln ( 2 )} + 1

Solution steps

2^{p - 1} + 5 = 3^{q}

Move

5

to the right side

2^{p - 1} = 3^{q} - 5

Apply exponent rules

(p - 1) ln (2) = ln (3^{q} - 5)

Solve

(p - 1) ln (2) = ln (3^{q} - 5) : p = \frac{ln ( 3 ^{q} - 5 )}{ln ( 2 )} + 1

p = \frac{ln ( 3 ^{q} - 5 )}{ln ( 2 )} + 1

e^{x} = 10

8^{x} = 2

(\frac{9}{4})^{2 x} = \frac{81}{16}

1 6^{x^{2}} = 8^{4 x - 3}

12 5^{x} = \frac{1}{25}

What is the answer to solvefor p,2^{p-1}+5=3^q ?
The answer to solvefor p,2^{p-1}+5=3^q is p=(ln(3^q-5))/(ln(2))+1