Question

find x so that (3)^x-2×3^5=3^3​

Answer 1

The value of x = 0

Given :-

3^(x-2) × 3^5 = 3^3

To find :-

The value of x

Solution :-

Given equation is 3^(x-2) × 3^5 = 3^3

We know that

a^m × a^n = a^(m+n)

Therefore, 3^(x-2+5) = 3^3

=> 3^(x+3) = 3^3

On comparing both sides then

=> x+3 = 3

=> x = 3-3

=> x = 0

Therefore, x = 0

Answer :-

The value of x is 0

Check :-

If x = 0 then LHS = 3^(x-2) × 3^5 becomes

3^(0-2)×3^5

=> 3^(-2) ×3^5

=> 3^(-2+5)

Since , a^m/a^n = a^(m-n)

=> 3^3

=> RHS

LHS = RHS is true for x = 0

Verified the given relations in the given problem.

Used formulae:-

♦ a^m × a^n = a^(m+n)

♦ a^m/a^n = a^(m-n)

Answer 2

Answer:

x=0

Step-by-step explanation:

Question:-

$\large{\sf{ {3}^{(x - 2)} \times {3}^{5} = {3}^{3} }}$

To find:-

• We have to find the value of x.

Given that:-

• We have been given that LHS =RHS.

Laws of exponent used:-

$\large\red{\sf{ {a}^{m} \times {a}^{n} = {a}^{(m + n)} }}$

Solution:-

$\large\underline{\sf{ {3}^{(x - 2)} \times {3}^{5} = {3}^{3} }}$

Using law of exponent we get:-

$\large\underline{\sf{ {3}^{(x -2) + 5} = {3}^{3} }}$

$\large\underline{\sf{ {3}^{x + 3} = {3}^{3} }}$

On ignoring base we get:-

$\large\underline{\sf{x + 3 = {3}}}$

$\large\underline{\sf{x = 3 - 3}}$

$\large\boxed{{ \boxed { \red{ \sf{ \implies x = 0}}}}}$

More laws of exponents:-

$\large \boxed{\sf{ \implies {a}^{m} \div {a}^{n} = {a}^{( m-n )} }}$

$\large\boxed{\sf{ \implies{a}^{0} = 1}}$

$\large\boxed{\sf{ \implies( \frac{ {a}^{m} }{ {b}^{m}}) = ( \frac{a}{b} )^{m} }}$

$\large\boxed{\sf{ \implies {a}^{m} \times {b}^{m} = {(ab)}^{m} }}$