What is the answer to solvefor a,(4^{a+1})/b =125 ?

The answer to solvefor a,(4^{a+1})/b =125 is a=(ln(125b))/(2ln(2))-1

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solvefor a,(4^{a+1})/b =125

solve for

a, \frac{4 ^{a + 1}}{b} = 125

a = \frac{ln ( 125 b )}{2 ln ( 2 )} - 1

Solution steps

\frac{4 ^{a + 1}}{b} = 125

Multiply both sides by

b

\frac{4 ^{a + 1}}{b} b = 125 b

Simplify

4^{a + 1} = 125 b

Apply exponent rules

(a + 1) ln (4) = ln (125 b)

Solve

(a + 1) ln (4) = ln (125 b) : a = \frac{ln ( 125 b )}{2 ln ( 2 )} - 1

a = \frac{ln ( 125 b )}{2 ln ( 2 )} - 1

8^{x} = 32

2^{n} = 2000

2 5^{x} = 0.25

3^{2 x + 2} = 9^{10}

1000 = 2^{n}

What is the answer to solvefor a,(4^{a+1})/b =125 ?
The answer to solvefor a,(4^{a+1})/b =125 is a=(ln(125b))/(2ln(2))-1