All categories

2 years ago

Find the value of y...

Mathematics

1 answer:

soldier1979 [14.2K]2 years ago

5 0

Answer:

y=0

Step-by-step explanation:

Rewrite Evaluate Powers:

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

Calculate:

$(3^{y})^{2}*\frac{1}{3} +2*3^{y}*\frac{1}{3}=1$

Solve using substitution:

$(3^{y})^{2}*\frac{1}{3} +\frac{2}{3} *3^{y}=1$ $t=3^{y}$

Solve the equation for t:

$t^{2}*\frac{1}{3}+\frac{2}{3}t=1$

t=-3

t=1

Substitute back to t=3^y

3^y=-3

3^y=1

y∉R

$3^{y}=1$

y=0

You might be interested in

Perform the following operations s and prove closure. Show your work.

nadezda [96]

Answer:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$ = $\frac{2x^2+10x+6}{(x+3)(x+5)}\\$

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$ = $\frac{1}{(x+2)(x-4)}$

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$ = $\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$

4. $\frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3}$ = $\frac{1}{(x-2)(x-4)}$

Step-by-step explanation:

1. $\frac{x}{x+3}+\frac{x+2}{x+5}$

Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)

$=\frac{x(x+5)+(x+2)(x+3)}{(x+3)(x+5)}\\=\frac{x^2+5x+(x)(x+3)+2(x+3)}{(x+3)(x+5)} \\=\frac{x^2+5x+x^2+3x+2x+6}{(x+3)(x+5)} \\=\frac{x^2+x^2+5x+3x+2x+6}{(x+3)(x+5)} \\=\frac{2x^2+10x+6}{(x+3)(x+5)}\\$

Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.

2. $\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}$

Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)

Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)

Putting factors

$=\frac{x+4}{(x+3)(x+2)}*\frac{x+3}{(x-4)(x+4)}\\\\=\frac{1}{(x+2)(x-4)}$

Prove closure: The value of x≠-2 and x≠4 because if there values are -2 and 4 then the denominator will be zero.

3. $\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}$

Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)

Factors of x^2-5x+6 = x^2-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)

Putting factors

$\frac{2}{(x+3)(x-3)}-\frac{3x}{(x+3)(x-2)}$

Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)

$\frac{2(x-2)-3x(x+3)(x-3)}{(x+3)(x-3)(x-2)}$

$=\frac{2(x-2)-3x(x+3)}{(x+3)(x-3)(x-2)}\\=\frac{2x-4-3x^2-9x}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-9x+2x-4}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}$