Transactional solutionpleasee#umarım yaparsınız

1.	Transactional solutionpleasee#umarım yaparsınız
Answer» 32 Step-by-step explanation: Given: $\sqrt[5]{27 {}^{(x - 2)} } = \sqrt[3] { {9}^{(x - 5)} }$ To find: The value of x Solution: #TABII ki SIZE yardımcı olacağım :) This question is related to Laws of Indices. So, We solve it by using the rules. 1st_Make the base exponent in smaller FORM i.e 25 = 5², 27 = 3³ 2nd_ Splitting the variable and constant powers i.e ${5}^{(x + 2)} = {5}^{x} . {5}^{2}$ Now, $\sqrt[5]{27 {}^{(x - 2)} } = \sqrt[3] { {9}^{(x - 5)} } \\ \\ \sqrt[5]{ {3}^{3(x - 2)} } = \sqrt[3]{ {3}^{2(x - 5)} } \\ \\ \sqrt[5]{ {3}^{(3x - 6)} } = \sqrt[3]{ {3}^{(2x - 10)} } \\ \\ { 3}^{ \{(3x - 6). \frac{1}{5} \} } ={ 3}^{ \{(2x - 10). \frac{1}{3} \} } \\ \\ {3}^{ \frac{3x}{5} } . {3}^{ \frac{ - 6}{5} } = {3}^{ \frac{2x}{3} } . {3}^{ \frac{ - 10}{3} } \\ \\ \frac{ {3}^{ \frac{ - 6}{5} } }{ {3}^{ \frac{ - 10}{3} } } = \frac{ {3}^{ \frac{2x}{3} } }{ {3}^{ \frac{3x}{5} } } \\ \\ {3}^{( \frac{ - 6}{5} + \frac{ 10}{3}) } = {3}^{( \frac{2x}{3} - \frac{3x}{5}) } \\ \\ {3}^{ (\frac{ - 6.3 + 10.5}{15} )} = 3 {}^{( \frac{5.2x - 3x.3}{15}) } \\ \\ [\text{Eliminating the base from both sides}] \\ \\ \frac{ - 18 + 50}{15} = \frac{10x - 9x}{15} \\ \\ \frac{ 32}{\cancel{15}} = \frac{x}{ \cancel{15}} \\ \\ x = 32$ Therefore, the required answer is 32. ${\underline{\blue{\text{ Some Important Laws of Indices}}}}$ ${a}^{n}.{a}^{m}={a}^{(n + m)}$ ${a}^{-1}=\dfrac{1}{a}$ $\dfrac{{a}^{n}}{ {a}^{m}}={a}^{(n-m)}$ ${({a}^{c})}^{b}={a}^{b\times c}={a}^{bc}$ ${a}^{\frac{1}{x}}=\sqrt[x]{a}$ [Where all variables are real and GREATER than 0] Varification See in the attachment.

Answer»

Step-by-step explanation:

Given:

$\sqrt[5]{27 {}^{(x - 2)} } = \sqrt[3] { {9}^{(x - 5)} }$

To find:

The value of x

Solution:

#TABII ki SIZE yardımcı olacağım :)

This question is related to Laws of Indices.

So, We solve it by using the rules.

1st_Make the base exponent in smaller FORM i.e 25 = 5², 27 = 3³

2nd_ Splitting the variable and constant powers

i.e

${5}^{(x + 2)} = {5}^{x} . {5}^{2}$

Now,

$\sqrt[5]{27 {}^{(x - 2)} } = \sqrt[3] { {9}^{(x - 5)} } \\ \\ \sqrt[5]{ {3}^{3(x - 2)} } = \sqrt[3]{ {3}^{2(x - 5)} } \\ \\ \sqrt[5]{ {3}^{(3x - 6)} } = \sqrt[3]{ {3}^{(2x - 10)} } \\ \\ { 3}^{ \{(3x - 6). \frac{1}{5} \} } ={ 3}^{ \{(2x - 10). \frac{1}{3} \} } \\ \\ {3}^{ \frac{3x}{5} } . {3}^{ \frac{ - 6}{5} } = {3}^{ \frac{2x}{3} } . {3}^{ \frac{ - 10}{3} } \\ \\ \frac{ {3}^{ \frac{ - 6}{5} } }{ {3}^{ \frac{ - 10}{3} } } = \frac{ {3}^{ \frac{2x}{3} } }{ {3}^{ \frac{3x}{5} } } \\ \\ {3}^{( \frac{ - 6}{5} + \frac{ 10}{3}) } = {3}^{( \frac{2x}{3} - \frac{3x}{5}) } \\ \\ {3}^{ (\frac{ - 6.3 + 10.5}{15} )} = 3 {}^{( \frac{5.2x - 3x.3}{15}) } \\ \\ [\text{Eliminating the base from both sides}] \\ \\ \frac{ - 18 + 50}{15} = \frac{10x - 9x}{15} \\ \\ \frac{ 32}{\cancel{15}} = \frac{x}{ \cancel{15}} \\ \\ x = 32$