Question

2^{y} + 3 ^{y} = 17

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

Solve for x and y

Answer 1

Is the first question properly typed ?  is it y or x as the power of x ?

For the first question there are two possible sets of x and y.

$2^x+3^y=17\\\\17=8+9=2^3+3^2,\ \ \ x=3,\ \ and\ \ y=2\\.\ \ =16+1=2^4+3^0,\ \ \ x=4\ and\ y=0$


There are at least two solutions to the second question.  Perhaps there are more solutions.

$2^{x+2}-3^{y-1}=5\\\\5=8-3=2^3-3^1,\\=\ \textgreater \ x+2=3,\ x=1,\ \ and\ \ y-1=1,\ \ y=2\\\\5=32-27=2^5-3^3\\=\ \textgreater \ x+2=5,\ \ x=3,\ \ \ and\ \ \ y-1=3,\ \ y=4$

Perhaps there is a general formula possible to derive.
==================
Let
$2^{a+1}-3^{b+1}=5=2+3\\2^{a+1}-2=3^{b+1}+3\\2(2^a-1)=3(3^b+1)\\\\So\ 2^a-1\ is\ multiple\ of\ 3\ and\ 3^b+1\ is\ a\ multiple\ of\ 2.$

Another way Let:
$2^{a+3}-3^{b+1}=5=2^3-3\\2^{a+3}-2^3=3^{b+1}-3\\2^3(2^a-1)=3(3^b-1)\\\\So\ 2^a-1\ is\ a\ multiple\ of\ 3.\ and\ 3^b-1\ is\ a\ multiple\ of\ 8.$

we have to find a and b such that the above relations are  valid.

Answer 2

Answer:

Is the first question properly typed ? is it y or x as the power of x ?

For the first question there are two possible sets of x and y.

\begin{gathered}2^x+3^y=17\\\\17=8+9=2^3+3^2,\ \ \ x=3,\ \ and\ \ y=2\\.\ \ =16+1=2^4+3^0,\ \ \ x=4\ and\ y=0\end{gathered}

2

x

+3

y

=17

17=8+9=2

3

+3

2

, x=3, and y=2

. =16+1=2

4

+3

0

, x=4 and y=0

There are at least two solutions to the second question. Perhaps there are more solutions.

\begin{gathered}2^{x+2}-3^{y-1}=5\\\\5=8-3=2^3-3^1,\\=\ \textgreater \ x+2=3,\ x=1,\ \ and\ \ y-1=1,\ \ y=2\\\\5=32-27=2^5-3^3\\=\ \textgreater \ x+2=5,\ \ x=3,\ \ \ and\ \ \ y-1=3,\ \ y=4\end{gathered}

2

x+2

−3

y−1

=5

5=8−3=2

3

−3

1

,

= \textgreater x+2=3, x=1, and y−1=1, y=2

5=32−27=2

5

−3

3

= \textgreater x+2=5, x=3, and y−1=3, y=4

Perhaps there is a general formula possible to derive.

==================

Let

\begin{gathered}2^{a+1}-3^{b+1}=5=2+3\\2^{a+1}-2=3^{b+1}+3\\2(2^a-1)=3(3^b+1)\\\\So\ 2^a-1\ is\ multiple\ of\ 3\ and\ 3^b+1\ is\ a\ multiple\ of\ 2.\end{gathered}

2

a+1

−3

b+1

=5=2+3

2

a+1

−2=3

b+1

+3

2(2

a

−1)=3(3

b

+1)

So 2

a

−1 is multiple of 3 and 3

b

+1 is a multiple of 2.

Another way Let:

\begin{gathered}2^{a+3}-3^{b+1}=5=2^3-3\\2^{a+3}-2^3=3^{b+1}-3\\2^3(2^a-1)=3(3^b-1)\\\\So\ 2^a-1\ is\ a\ multiple\ of\ 3.\ and\ 3^b-1\ is\ a\ multiple\ of\ 8.\end{gathered}

2

a+3

−3

b+1

=5=2

3

−3

2

a+3

−2

3

=3

b+1

−3

2

3

(2

a

−1)=3(3

b

−1)

So 2

a

−1 is a multiple of 3. and 3

b

−1 is a multiple of 8.

we have to find a and b such that the above relations are valid.

Please brainlest me the answer