Question
Solve: $2^x+3^y=17,2^{x+2}-3^{y+1}=5$
✓
Answer
$2^x+3^y=17...(i)$
$2^{x+2}-3^{y+1}=5$
$\therefore \quad 2^x \cdot 2^2-3^y \cdot 3^1=5$
$\therefore \quad 4.2^x-3.3^y=5 ...(ii)$
Substituting $2^x=m$ and $3^y=n$, we get,
$m+n=17...(iii)$
$4 m-3 n=5...(iv)$
Multiplying eq. $(iii)$ by $3,$
$ 3m + 3n = 51...(v)$
Adding $(iv)$ and $(v)$ we get,
$\therefore \quad m=\frac{56}{7}$
$\therefore \quad m=8$
Substituting $m=8$ in $(iii),$
$8+n=17$
$\therefore \quad n=17-8$
$\therefore \quad n=9$
Resubstituting the value of $m$ and $n$, we get,
$ m=2^x$
$\therefore 8=2^x$
$\therefore 2^3=2^x$
$\therefore x=3$
$n=3^y$
$9=3^y$
$3^2=3^y$
$y=2$
$\therefore x=3$ and $y=2$ is the solution.
$2^{x+2}-3^{y+1}=5$
$\therefore \quad 2^x \cdot 2^2-3^y \cdot 3^1=5$
$\therefore \quad 4.2^x-3.3^y=5 ...(ii)$
Substituting $2^x=m$ and $3^y=n$, we get,
$m+n=17...(iii)$
$4 m-3 n=5...(iv)$
Multiplying eq. $(iii)$ by $3,$
$ 3m + 3n = 51...(v)$
Adding $(iv)$ and $(v)$ we get,
$\therefore \quad m=\frac{56}{7}$
$\therefore \quad m=8$
Substituting $m=8$ in $(iii),$
$8+n=17$
$\therefore \quad n=17-8$
$\therefore \quad n=9$
Resubstituting the value of $m$ and $n$, we get,
$ m=2^x$
$\therefore 8=2^x$
$\therefore 2^3=2^x$
$\therefore x=3$
$n=3^y$
$9=3^y$
$3^2=3^y$
$y=2$
$\therefore x=3$ and $y=2$ is the solution.
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