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Given, 13x + 45 = - 20

13x + 45 = - 20

Subtract 45 from each side

=> 13x + 45 - 45 = - 20 - 45

=> 13x = - 65

Step 2:

Divide each side by 13

=> $\frac{13x}{13} = \frac{- 65}{13}$

=> x = - 5.

Given log $\frac{xy^3}{z^2}$

Step 1:

log $\frac{xy^3}{z^2}$

=> log xy^{3} - log z^{2}

[log$\frac{m}{n}$ = log m - log n]

Step 2:

Again

log xy^{3} - log z^{2} = log x + log y^{3} - log z^{2}

[log mn = log m + log n]

Step 3:

log x + log y^{3} - log z^{2} = log x + 3 log y - 2 log z

[log m^{n} = n log m]

=> $\frac{xy^3}{z^2}$ = log x + 3 log y - 2 log z.**answer**

Step 1:

log $\frac{xy^3}{z^2}$

=> log xy

[log$\frac{m}{n}$ = log m - log n]

Step 2:

Again

log xy

[log mn = log m + log n]

Step 3:

log x + log y

[log m

=> $\frac{xy^3}{z^2}$ = log x + 3 log y - 2 log z.

2x

Given quadratic equation, 2x^{2} + 7x - 4 = 0

Step 1:

2x^{2} + 7x - 4 = 0

Comparing with general quadratic equation

ax^{2} + bx + c = 0

a = 2, b = 7, c = - 4

**Step 2:**

b^{2} - 4ac = (7)^{2 }- 4 * 2 * - 4

= 49 + 32

= 81

and

$\sqrt{b^2 - 4ac}$ = $\sqrt{81}$

= 9

=> $\sqrt{b^2 - 4ac}$ = 9

Step 3:

x = $\frac{- b \pm\sqrt{b^2 - 4ac}}{2a}$

=> x = $\frac{- 7 \pm9}{2*2}$

=> x = $\frac{- 7 + 9}{4}$ and x = $\frac{- 7 - 9}{4}$

=> x = $\frac{2}{4}$ and x = $\frac{- 16}{4}$

=> x = $\frac{1}{2}$ and x = - 4

Hence the values of x are: $\frac{1}{2}$ , - 4.**answer**

Step 1:

2x

Comparing with general quadratic equation

ax

a = 2, b = 7, c = - 4

b

= 49 + 32

= 81

and

$\sqrt{b^2 - 4ac}$ = $\sqrt{81}$

= 9

=> $\sqrt{b^2 - 4ac}$ = 9

Step 3:

x = $\frac{- b \pm\sqrt{b^2 - 4ac}}{2a}$

=> x = $\frac{- 7 \pm9}{2*2}$

=> x = $\frac{- 7 + 9}{4}$ and x = $\frac{- 7 - 9}{4}$

=> x = $\frac{2}{4}$ and x = $\frac{- 16}{4}$

=> x = $\frac{1}{2}$ and x = - 4

Hence the values of x are: $\frac{1}{2}$ , - 4.

Given 2x

2x

=> 2x

=> 2x

=> 2x

Step 2:

To find the value of p, put x = 1

=> 2 * 1

=> 2 * 1 + 2p - p = 4

=> 2 + p = 4

Subtract 2 from both sides

=> 2 + p - 2 = 4 - 2

=> p = 2

Given, $\frac{1}{m + 1} + \frac{2}{2m + 3}$

Step 1:

$\frac{1}{m + 1} + \frac{2}{2m + 3}$

LCM of the fractions = (m + 1)(2m + 3)

Step 2:

$\frac{1}{m + 1} + \frac{2}{2m + 3}$

= $\frac{2m + 3 + 2(m + 1)}{(m + 1)(2m + 3)}$

= $\frac{2m + 3 + 2m + 2}{(m + 1)(2m + 3)}$

= $\frac{4m + 5}{(m + 1)(2m + 3)}$

=> $\frac{1}{m + 1} + \frac{2}{2m + 3}$ = $\frac{4m + 5}{(m + 1)(2m + 3)}$

Step 1

$\frac{1}{m + 1} + \frac{2}{2m + 3}$

LCM of the fractions = (m + 1)(2m + 3)

Step 2:

$\frac{1}{m + 1} + \frac{2}{2m + 3}$

= $\frac{2m + 3 + 2(m + 1)}{(m + 1)(2m + 3)}$

= $\frac{2m + 3 + 2m + 2}{(m + 1)(2m + 3)}$

= $\frac{4m + 5}{(m + 1)(2m + 3)}$

=> $\frac{1}{m + 1} + \frac{2}{2m + 3}$ = $\frac{4m + 5}{(m + 1)(2m + 3)}$