Question

help friend ques 7...

Answer 1

HELLO DEAR,

$\frac{ {7}^{3} }{ {7}^{(x - 2)} } = {7}^{7} \\ \\ \\ = {7}^{3} = {7}^{7} \times {7}^{x - 2} \\ \\ \\ = \frac{ 7 \times 7 \times 7 \times }{7 \times 7 } = {7}^{x - 2} \\ \\ \\ = > 7 = {7}^{x - 2} \\ \\ \\ = 1 = x - 2 \\ \\ \\ = x = 2 + 1 \\ \\ \\ = x = 3$

second question,

9^x ÷ 9^-5 = 9^13

=> 9^2x × 1/9^-5 = 9^13

=> 9^2x × 9^5 = 9^13

=> 9^2x = 9^13/9^5

=> 9^2x = 9^8

HENCE ,

2x = 8

=> x = 4

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Heya user !!

Let me tell you few important Formula that I will

be using over here

$\frac{ {a}^{m} }{ {a}^{n \: } } \: = {a}^{m - n}$
Now coming to our Question -

$( \: i \: ) \: \: \frac{ {7}^{3} }{ {7}^{x - 2} } = {7}^{7} \\ \\ = {7}^{3 - (x - 2)} = {7}^{7} \\ \\ = {7}^{3 - x + 2} \: = {7}^{7} \\ \\ = - x + 5 \: = 7 \\ \\ x = - 2$
Se know if the bases are the Same then its power are added


Now , Coming to the 2nd Question

$ii) \: \frac{ {9}^{2x} }{ {9}^{ - 5} } = {9}^{13 } \\ \\ = {9}^{2x + 5} = {9}^{13} \\ \\ = > x = 4$


You may verify it by substituting the value of x

Thanks