Answer:
ai) [tex]m^3[/tex]
aii) 2
Step-by-step explanation:
ai)
When multiplying variables with same base, you add the exponents.
[tex]m^{-5} \cdot m^{8}=m^{-5+8}=m^3[/tex]
aii)
[tex]3^{2+n}=81[/tex]
We can do this with logarithms or without.
I will do both.
Without logarithms:
I'm going to write both sides so they have the same base which is 3.
We can do that since [tex]3=3^1[/tex] and [tex]81=3^4[/tex].
This gives us:
[tex]3^{2+n}=3^{4}[/tex]
Nothing about the left hand side change because it was already written with base 3 and also if you do 1(2+n) you still get 2+n.
Now that the bases are the same we can set the exponents equal:
2+n=4
Subtract 2 on both sides:
n=4-2
Simplify:
n=2
So n=2 renders the equation true.
Check:
[tex]3^{2+n}=81[/tex] with n=2
[tex]3^{2+2}=81[/tex]
[tex]3^4=81[/tex]
[tex]81=81[/tex]
That is a true equation so n=2 is indeed a solution.
Now let's do it the logarithm way.
To do it this way you want the exponential expression by itself without any constant multiple except the exception of 1. We have that already so we are ready to take log of both sides:
[tex]3^{2+n}=81[/tex]
Take log( ) of both sides:
[tex]\log(3^{2+n})=\log(81)[/tex]
This allows us to bring the variable exponent expression down by use of power rule.
[tex](2+n)\log(3)=\log(81)[/tex]
Divide boht sides by [tex]\log(3)[/tex]:
[tex]2+n=\frac{\log(81)}{\log(3)}[/tex]
If you input right hand side into calculator it should give you 4:
[tex]2+n=4[/tex]
Subtract 2 on both sides:
[tex]n=4-2[/tex]
Simplify:
[tex]n=2[/tex]
My favorite way is the first when you can do it.
