Question

2^x+3^y=17;2^x+2-3^y-1=5

Answer 1

Given:

$2^{x} + 3^{y} = 17$            → I

$2^{x+2} - 3^{y-1} =5$       → II

To Find:

x and y

Solution:

We must recall the following laws of indices to solve this question,

1. $x^{m} * x^{n} = x^{m+n}$
2. $x^{m} / x^{n} = x^{m-n}$

Now let,

$2^{x}$ ⇒ A

$3^{y}$ ⇒ B

Equation I : $2^{x} + 3^{y} = 17$

                   Therefore,

                   A + B = 17

Use the above laws to simplify equation II

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

We get:

$2^{x+2} = 2^{x} * 2^{2}\\3^{y-1} = 3^{y} / 3^{1}$

$(2^{x} * 2^{2}) - (3^{y} / 3^{1} ) = 5$

Replace  $3^{y}$ ⇒ B

               $2^{x}$ ⇒ A

⇒ $(A*2^{2}) - (B/3^{1}) = 5$

⇒ $4A - \frac{1}{3}B = 5$

We have now acquired two new equations:

$A+B=17$                 →       Eqn. a

$4A - \frac{1}{3}B = 5$               →       Eqn. b

Use the method of substitution to find the values of the A & B variables.

(here, we are making A the subject from eqn a to substitute into eqn b)

⇒ $A+B=17$

⇒ $A=B-17$       →  III

III in eqn b.

$4(B-17) - \frac{1}{3}B = 5$

Solve;

$4(17-B) - \frac{1}{3}B = 5\\\\3[4(17-B) - \frac{1}{3}B] = (5)3\\\\12(17-B) - B = 15\\\\204 -12B -B=15\\\\204-15=12B+B\\\\189 = 13B\\\\B = 189/13$

Put the value of B in eqn a.

$A +\frac{189}{13} =17\\\\A=17 - \frac{189}{13}\\\\A= \frac{32}{13}$

Finally use the logarithm function to find x and y;

$2^{x}$ ⇒ A ⇒ $\frac{32}{13}$

$3^{y}$ ⇒ B ⇒ $\frac{189}{13}$

Follow these logarithmic rules:

1. logₐ $x^{m}$ = m logₐ x
2. logₐ $\frac{m}{n}$ = logₐ m - logₐ n
3. logₐ mn = logₐ m + logₐ n

x log2 = log $\frac{32}{13}$

x log2 = log 32 - log 13

x = (log 32 - log 13) / log 2

y log3 = log $\frac{189}{13}$

y log3 = log 189 - log 13

y = (log 189 - log 13) / log 3

Hence, x = (log 32 - log 13) / log 2 and y = (log 189 - log 13) / log 3

Answer 2

Given,

$2^{x}+3^{y}=17$ and $2^{x+2}-3^{y-1}=5$

We have to find the values of $x$ and $y$.

Formulas:

$x^{m}*x^{n}=x^{m+n}$

$x^{m}/x^{n}=x^{m-n}$

Let, $2^{x}=A$ and $3^{y}=B$

The equation $2^{x}+3^{y}=17$ becomes $A+B=17$

$A+B=17$ .......($1$)

Use the above formulas, to simplify the equation $2^{x+2}-3^{y-1}=5$

$2^{x+2}=2^{x}*2^{2}$

$3^{y-1}=\frac{3^{y} }{3^{1} }$

Substitute the above values in $2^{x+2}-3^{y-1}=5$

$2^{x}*2^{2}-\frac{3^{y} }{3^{1}}=5$

Replace $2^{x}=A$ and $3^{y}=B$

$4A-\frac{B}{3}=5$

$12A-B=15$ .....($2$)

Adding the both equations, we get

$A+B+12A-B=17+15$

$13A=32$

$A=\frac{32}{13}$

Now substitute $A=\frac{32}{13}$ in equation ($1$),we get

$\frac{32}{13}+B=17$

$32+13B=221$

$B=\frac{189}{13}$

We know that,

$2^{x}=A$

$3^{y}=B$

Now, we can use the logarithm functions to find the values of $x$ and $y$.

$log2^{x}=x log 2$

$xlog2=log\frac{32}{13}$

$x=\frac{log32-log13}{log2}$

$log3^{y}=ylog3$

$ylog3=log\frac{189}{13}$

$y=\frac{log189-log13}{log3}$

Therefore, the values of $x$ and $y$ are respectively $\frac{log32-log13}{log2}$ and $\frac{log189-log13}{log3}$.

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