Test: Exponents- 3 - Mechanical Engineering MCQ

Putting these together,

Then putting them together,  

The easiest way to simplify is to work from the inside out. We should first get rid of the negatives in the exponents. Remember that variables with negative exponents are equal to the inverse of the expression with the opposite sign. For example
  So using this, we simplify:  

Now when we multiply variables with exponents, to combine them, we add the exponents together. For example,  (a²)(a²)=a²⁺²=a⁴ Doing this to our expression we get it simplified to 
The next step is taking the inside expression and exponentiating it. When taking an exponent of a variable with an exponent, we actually multiply the exponents. For example,  (a²)⁴=a^(2∗4)=a⁸.
The other rule we must know that is an exponent of one half is the same as taking the square root. So for the So using these rules,

First, write each expression as a base 3 logarithm:
3=log₃27 since 3³=27
2log₃y=log₃y²
Rewrite the expression accordingly, and apply the logarithm sum and difference rules:
3+log₃x−2log₃y
=log₃27+log₃x−log₃y²

=log₃(27⋅x)−log₃y²

We have 3^p+3^p+3^p=3∗3^p=3^p+1=3q.

So p+1=q, and p=q−1.

Inspect the first few powers of 6; a pattern emerges.

6¹=6

6²=36

6³=216

6⁴=1,296

6⁵=7,776

6⁶=46,656

6⁷=279,936

6⁸=1,679,616

6⁹=10,077,696

6¹⁰=60,466,176

As you can see, the last two digits repeat in a cycle of 5. 

789 divided by 5 yields a remainder of 4; the pattern that becomes apparent in the above list is that if the exponent divided by 5 yields a remainder of 4, then the power ends in the diigts 96.

Use the properties of exponents as follows:

(x³)⁴⋅(x³)⁵

=x^3⋅4 ⋅ x^3⋅5

=x¹²⋅ x¹⁵

=x¹²⁺¹⁵

=x²⁷

Apply the power of a power principle twice by multiplying exponents:

(1/25)^N ⋅ 5⁶=125

(5⁻²)^N ⋅ 5⁶=53

5^−2⋅N ⋅ 5⁶=53

5^−2N + 6=53

−2N+6=3

−2N+6−6=3−6

−2N=−3

−2N÷(−2)=−3÷(−2)

N=3/2

log8+log5−log4
=log(8⋅5)−log4
=log40−log4
=log(40÷4)
=log10=1

4log3−2log6

=log(3⁴)−log(6²)

=log81−log36

=log81/36

=log9/4

=log 9− log 4

5⁴=625;5⁵=3,125

Therefore, 

log₅ 625=4;log⁵ 3,125=5

log₅625<log₅1,000<log₅3,125

4<log₅1,000<5