Question

using laws of exponents verify the following statement are true b) 9^2n-1=1/9(81)^n​

Answer 1

Step-by-step explanation:

$\green{\mathbb{\red{GIVEN :{9}^{2n-1}=\frac{1}{9}\times {81}^{n}}}}$

1 Simplify $\sf\dfrac{1}{9}\times {81}^{n}$ to $\sf\dfrac{{81}^{n}}{9}$

$\tt \purple{ \implies{9}^{2n-1}=\dfrac{{81}^{n}}{9}}$

2 Multiply both sides by 99.

$\tt \pink{ \implies{9}^{2n}={81}^{n}}$

3 Convert both sides to the same base.

$\tt \red{ \implies{9}^{2n}={9}^{2n}}$

$\therefore$ following statement are true.

4 Since both sides equal, there are infinitely many solutions.

Infinitely Many Solutions.