What is the answer to solvefor x,y-3=4(5^{2x+1}) ?

The answer to solvefor x,y-3=4(5^{2x+1}) is x=(ln((y-3)/4)-ln(5))/(2ln(5))

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solvefor x,y-3=4(5^{2x+1})

solve for

x, y - 3 = 4 (5^{2 x + 1})

x = \frac{ln ( \frac{y - 3}{4} ) - ln ( 5 )}{2 ln ( 5 )}

Solution steps

y - 3 = 4 (5^{2 x + 1})

Switch sides

4 \cdot 5^{2 x + 1} = y - 3

Divide both sides by

4

5^{2 x + 1} = \frac{y - 3}{4}

Apply exponent rules

(2 x + 1) ln (5) = ln (\frac{y - 3}{4})

Solve

(2 x + 1) ln (5) = ln (\frac{y - 3}{4}) : x = \frac{ln ( \frac{y - 3}{4} ) - ln ( 5 )}{2 ln ( 5 )}

x = \frac{ln ( \frac{y - 3}{4} ) - ln ( 5 )}{2 ln ( 5 )}

57 = 1 3^{x}

2^{6 \cdot l o g_{4} (x)} = 8

9^{x + 12} = 8 1^{x + 5}

e^{3 - 4 x} = 8

e^{3 - 4 x} = 6

What is the answer to solvefor x,y-3=4(5^{2x+1}) ?
The answer to solvefor x,y-3=4(5^{2x+1}) is x=(ln((y-3)/4)-ln(5))/(2ln(5))