How do you solve ${\sqrt{7}}^{6 x} = 49^{x - 6}$ $sqrt7^(6x)= 49^(x-6)$ ?

Question

1599 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-14T06:18:37

$x = - 12$ $x = -12$

Replace $\sqrt{7}$ $sqrt7$ by $7^{\frac{1}{2}}$ $7^(1/2)$ and 49 by $7^{2} .$ $7^2.$ .

Now, use ${(a^{m})}^{n} = a^{m n}$ $(a^m)^n=a^(mn)$ .

$7^{3 x} = 7^{2 x - 12}$ $7^(3x)=7^(2x-12)$ , Compare exponents..

$3 x = 2 x - 12$ $3x=2x-12$ .

$x = - 12$ $x=-12$ .

How do you solve √76x=49x−6sqrt7^(6x)= 49^(x-6)?