Question 707147
{{{ 2(x-1)^(3/4)+4=36 }}}
To solve for x wed will "peel away" everything but the x from the left side. First we get rid of the extra term by subtracting 4 from each side:
{{{ 2(x-1)^(3/4)=32 }}}
Next we'll eliminate the 2 by dividing both sides by 2:
{{{ (x-1)^(3/4)=16 }}}
Next we eliminate the exponent. This is a little easier than it may seem at first if you realize that you're not actually eliminating the exponent. <i>Every number/expression</i> has an exponent! It's just that if the exponent is a 1 we usually don't bother writing it. So what we are really doing is changing the exponent into a 1. How can we change an exponent from 3/4 to 1? How do we change exponents in any way? We have several rules which tell us how exponents can be changed properly. The fastest, easiest way to change the exponent from 3/4 into a 1 is to use the power of a power rule. We're going to raise both sides of the equation to whatever power which will end up turning that 3/4 into a 1. The power of a power rule says that we should multiply the exponents. So 3/4 times what is 1? If you understand reciprocals you know the answer. Multiplying reciprocals <i>always</i> results in a 1! So we will raise each side of the equation by the reciprocal of 3/4 (which is 4/3):
{{{ ((x-1)^(3/4))^(4/3)=16^(4/3) }}}
On the left we get the exponent of 1 we were looking for:
{{{ x-1 = 16^(4/3) }}}
The only thing left to eliminate on the left side is the 1. Adding 1 to each side:
{{{ x = 1+16^(4/3) }}}
This may be an acceptable answer. But perhaps you (or your teacher) would prefer the answer in simplified radical form (since the fractional exponent on the right indicates a root of some kind). First let's write the expression with a radical. In general {{{x^(m/n) = root(n, x^m) = (root(n, x))^m}}}. I'm going to use the first form:
{{{ x = 1+root(3, 16^4) }}}
This radical will simplify if there are any perfect cube factors in {{{16^4}}}. Since 8 is a perfect cube and since 8 is a factor of 16 we definitely have perfect cube factors in {{{16^4}}}:
{{{ x = 1+root(3, (8*2)^4) }}}
{{{ x = 1+root(3, (8*2)*(8*2)*(8*2)*(8*2)) }}}
{{{ x = 1+root(3, 8*8*8*8*(2*2*2)*2) }}}
{{{ x = 1+root(3, 8*8*8*8*8*2) }}}
Now we can use a property of radicals, {{{root(n, a*b) = root(n, a)*root(n, b)}}}, to separate each of the factors into its own cube root:
{{{ x = 1+root(3, 8)*root(3, 8)*root(3, 8)*root(3, 8)*root(3, 8)*root(3, 2) }}}
Each of the cube roots of the perfect cubes will simplify:
{{{ x = 1+2*2*2*2*2*root(3, 2) }}}
which simplifies to:
[Correction: 2*2*2*2*2 is 32 (as the other tutor showed) not 16 as in my initial solution.]
{{{ x = 1+32*root(3, 2) }}}
This is an exact expression for the solution in simplified radical form.