What is 1/(2^x)+2/(3^y)=10, 3/(2^x)+7/(3^y)=33 ?

The solution to 1/(2^x)+2/(3^y)=10, 3/(2^x)+7/(3^y)=33 is

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1/(2^x)+2/(3^y)=10, 3/(2^x)+7/(3^y)=33

\frac{1}{2 ^{x}} + \frac{2}{3 ^{y}} = 10, \frac{3}{2 ^{x}} + \frac{7}{3 ^{y}} = 33

(x = - 2, y = - 1)

Solution steps

[\frac{1}{2 ^{x}} + \frac{2}{3 ^{y}} = 10 \frac{3}{2 ^{x}} + \frac{7}{3 ^{y}} = 33]

Isolate

3^{y}

for

\frac{1}{2 ^{x}} + \frac{2}{3 ^{y}} = 10 : 3^{y} = - \frac{2 ^{1 + x}}{1 - 5 \cdot 2 ^{1 + x}}; x \neq = - \frac{ln ( 5 )}{ln ( 2 )} - 1

Plug the solutions

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

into

\frac{3}{2 ^{x}} + \frac{7}{3 ^{y}} = 33

For

\frac{3}{2 ^{x}} + \frac{7}{3 ^{y}} = 33

, subsitute

3^{y}

with

- \frac{2 ^{1 + x}}{1 - 5 \cdot 2 ^{1 + x}} : x = - 2

Plug the solutions

x = - 2

into

\frac{1}{2 ^{x}} + \frac{2}{3 ^{y}} = 10

For

\frac{1}{2 ^{x}} + \frac{2}{3 ^{y}} = 10

, subsitute

x

with

- 2 : y = - 1

Verify solutions by plugging them into the original equations

Therefore, the final solution for

\frac{1}{2 ^{x}} + \frac{2}{3 ^{y}} = 10, \frac{3}{2 ^{x}} + \frac{7}{3 ^{y}} = 33

(x = - 2, y = - 1)

a + b = 27, a \cdot b = 182

x (n) = 1 + 3 x (\frac{n}{2}), x (1) = 0

x + y = 3, x^{3} + y^{3} = 9

\frac{m}{s ^{2}} v = 3.2 \frac{m}{s}, a = 0

7^{y} = 25 5^{z}, 1 5^{x} = 1

What is 1/(2^x)+2/(3^y)=10, 3/(2^x)+7/(3^y)=33 ?
The solution to 1/(2^x)+2/(3^y)=10, 3/(2^x)+7/(3^y)=33 is