Question 571240
<pre>
{{{(1/16)^(-2v)}}}·64 = 1

Learn the rule:  "Flip the fraction, change the sign of the exponent":
That is, {{{(A/B)^(-C)}}} becomes {{{(B/A)^C}}}

{{{(1/16)^(-2v)}}}·64 = 1

becomes

{{{(16/1)^(2v)}}}·64 = 1

And we can dispense with the 1 denominator under the 16

   16<sup>2v</sup>·64 = 1

Since 16 = 2·2·2·2 = 2<sup>4</sup> and since 64 = 2·2·2·2·2·2 = 2<sup>6</sup>

(2<sup>4</sup>)<sup>2v</sup>·2<sup>6</sup> = 1

Remove the parentheses by multiplying the exponent inside 4
by the exponent outside 2v, getting an exponent of 8v

2<sup>8v</sup>·2<sup>6</sup> = 1

Add the exponents

2<sup>8v+6</sup> = 1

We know that 0 is the only exponent 2 can have to give 1,
since 2<sup>0</sup> = 1, so that means 8v+6 must equal 0.  So

8v + 6 = 0

    8v = -6

     v = {{{-3/4}}} 

Edwin</pre>