6 Answers | Add Yours
1) "Roots" in this context are x values so in the first question n=4, and x=81, we are looking for the simplified version of `root(4)(81)` which, for the purposes of understanding can be written as:
`root(4)(9 times 9) = root(4)(3 times3times3times3)`
and as we have root 4 and we have four 3s, effectively the exponent (4) cancels out the root (4):
`therefore = 3`
2) When n=5 and `x=x^5` , we have `root(5)(x^5)` we can apply the same principle. Just as two of anything cancels a square root, 3 of anything cancels a cubed root and so on, we have a root 5 and we have five x-es:
`root(5)(x^5)= root(5)(x times x times x times x times x)`
3) To solve:
we convert it using an exponent because, logs and exponents go together. Therefore it becomes:
Note how the log base (8) becomes the exponential base (8). Now simplify because 8 is `2^3` and 32 is `2^5` such that:
`(2^3)^x = 2^5`
`therefore 2^(3x) = 2^5`
As the bases are now the same (ie 2), we can create an equation from the powers:
4) Solve this one the same way but there is no need to simplify:
`log_(10)(x) = 1`
1) `root(4)(81) = 3`
2) `root(5)(x^5)= x`
3) `x= 4/3`
Everything as durbanville had except #1, you also need to include -3.
And, for #3, x = 5/3. There was a typo at:
and 32 is such that:
At the second line, (2^3)^x = 2^5.
The first thing you have to know about these problems is that roots can be written as exponential fractions and the change of base rules for logs.
1) First know that the forth root of any # or variable can be written as x^1/4, and then plugged into the calculator. But if need be, its the same as taking the square root of the # twice, in this case the fourth root of 81 is 3.
2) For simplicity, you can say that your multiplying fifth root of x, which is x^1/5, by x^5. Just multiply 1/5 by 5 and get x, they cancel each other out to become 1, which just leaves you with the variable.
3) Logb x = N, is the same as b^N = x. So log8 32 = x is the same as 8^x = 32, Do a little calculator magic, and x = 5/3.
4) Same as previous problem: log10 x =1, 10^1 = x, x=10.
1) The key to solve problems 1 and 2 is to look at the quantity underneath the root sign and try to factorize it in such a way that the calculation becomes simpler.
For example, 81 = 9 x 9 = 3 x 3 x 3 x 3 = 3^4
And we are aiming to find 4th root, means to the power 1/4
So, the required answer = 3^[4*(1/4)]
= 3^1 = 3
Similarly, this holds true for -3
2) Similarly, here the answer = x^[5*(1/5)] = x
3) In problems 3 and 4, you have to look at the quantities and simplify it in such a manner that the calculation becomes easier.
For example, log(32) to the base 8 = x
or 8^x = 32
or 8^x = 2^5
or 2^3x = 2^5
or 3x = 5
or x = 5/3
4) Similarly, log(x) = 1
or 10^1 = x
or x = 10
These problems can be easily solved with a graphing calculator. However, if you don't have one, rewriting them could make them easier to understand.
1. `root(4)(81)` can be rewritten as `x^(4)=81`. All you have to do is think what can be multiplied by itself 4 times in order to get 81. The answer is `+-3` (negatives are included because when a negative is multiplied by another negative you get a positive number. Since it's being multiplied an even number of times, you will end up with a positive 81 either way).
2. `root(5)(x^5)` is easier than it looks to solve. Because you have a square root 5 and an exponent of 5, these simply cancel out and you are left with x.
3. log8(32)=x can also be rewritten where the base (8) has an exponent of the answer (x). So, log8(32)=x is equivalent to `8^(x)=32`. You can now simplify the bases which gives you `(2^3)^(x)=(2^5)`.
`2^(3x)=(2^5)` The bases cancel
4. log10(x)=1 rewritten would be `10^(1)=x` which is simply x=10.
durbanville...has already answered correct answer.
We’ve answered 318,911 questions. We can answer yours, too.Ask a question