Question

) If 5 = 3, find the value of −.
+

Answer 1

Given, 5

Given, 5 3x−1

Given, 5 3x−1 ÷25=125

Given, 5 3x−1 ÷25=125Since, 25=5×5=5

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5)

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5)

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5)

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5)

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a m

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a m ÷a

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a m ÷a n

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a m ÷a n =(a)

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a m ÷a n =(a) m−n

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a m ÷a n =(a) m−n ]

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a m ÷a n =(a) m−n ]⇒(5)

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a m ÷a n =(a) m−n ]⇒(5) 3x−3

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a m ÷a n =(a) m−n ]⇒(5) 3x−3 =(5)

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a m ÷a n =(a) m−n ]⇒(5) 3x−3 =(5) 3

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a m ÷a n =(a) m−n ]⇒(5) 3x−3 =(5) 3

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a m ÷a n =(a) m−n ]⇒(5) 3x−3 =(5) 3 On comparing both sides, we get

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a m ÷a n =(a) m−n ]⇒(5) 3x−3 =(5) 3 On comparing both sides, we get⇒3x−3=3

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a m ÷a n =(a) m−n ]⇒(5) 3x−3 =(5) 3 On comparing both sides, we get⇒3x−3=3⇒3x=6

Given, 5 3x−1 ÷25=125Since, 25=5×5=5 2 And 125=5×5×5=5 3 Therefore, 5 3x−1 ÷(5) 2 =(5) 3 ⇒(5) 3x−1−2 =(5) 3 [∵a m ÷a n =(a) m−n ]⇒(5) 3x−3 =(5) 3 On comparing both sides, we get⇒3x−3=3⇒3x=6⇒x=2