Question

(3/7)^4+(3/7)^5=(3/7)^2m+1. find the value of m

Answer 1

GIVEN :

The equation is $(\frac{3}{7})^4\times (\frac{3}{7})^5=(\frac{3}{7})^{2m+1}$

TO FIND :

The value of m in the given equation.

SOLUTION :

Given that the equation is $(\frac{3}{7})^4\times (\frac{3}{7})^5=(\frac{3}{7})^{2m+1}$

Now solving the given equation as below:

$(\frac{3}{7})^4\times (\frac{3}{7})^5=(\frac{3}{7})^{2m+1}$

By using the exponent property:

$a^m.b^m=a^{m+n}$

$(\frac{3}{7})^{4+5}=(\frac{3}{7})^{2m+1}$

$(\frac{3}{7})^{9}=(\frac{3}{7})^{2m+1}$

∵ the bases are equal so that we can equate the powers of the above equation we get,

9=2m+1

Rewritting the above equation we get,

2m+1=9

2m=9-1

2m=8

$m=\frac{8}{2}$

∴ m=4.

∴ the value of m in the given equation is 4.