What is 8^x=2^{y+1},5^y=25^{x+1} ?

The solution to 8^x=2^{y+1},5^y=25^{x+1} is

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8^x=2^{y+1},5^y=25^{x+1}

8^{x} = 2^{y + 1}, 5^{y} = 2 5^{x + 1}

(x = 3, y = 8)

Solution steps

[8^{x} = 2^{y + 1} 5^{y} = 2 5^{x + 1}]

Isolate

x

for

8^{x} = 2^{1 + y} : x = \frac{1 + y}{3}

Plug the solutions

x = \frac{1 + y}{3}

into

5^{y} = 2 5^{1 + x}

For

5^{y} = 2 5^{1 + x}

, subsitute

x

with

\frac{1 + y}{3} : y = 8

Plug the solutions

y = 8

into

8^{x} = 2^{1 + y}

For

8^{x} = 2^{1 + y}

, subsitute

y

with

8 : x = 3

Verify solutions by plugging them into the original equations

Therefore, the final solution for

8^{x} = 2^{y + 1}, 5^{y} = 2 5^{x + 1}

(x = 3, y = 8)

5 x_{1} + 7 x_{2} = 3, 2 x_{1} + 4 x_{2} = 1

2 y - x = - 9, 7 y - 2 x = 3

2 u + 3 v = 13, 4 u + 5 v = 23

3 x + 5 y = 38, 3 x - y = - 4

11 x + 5 y = 7, 6 x - 3 y = 21