How do you solve -3log5x=-6?
1 answer:
-3log5x = -6 would equal x = 25.
First, use the Power Rule: (log b x
= c log b x). Your problem should look like -log5 x³ = -6.
Second, multiply both sides by -1. Your problem should look like log5 x³ = 6. We removed the negative signs.
Third, use the Definition of Common Logarithms. ( b
= x if and only if logb(x) = a). Your problem should look like x³ = 5
.
Fourth, simplify 5 to 15625. Your problem should look like x³ = 15625.
Fifth, take the cube root of both sides. Your problem should look like x = 3√15625.
Sixth, calculate your problem to get x = 25 which is the answer.
First, use the Power Rule: (log b x
Second, multiply both sides by -1. Your problem should look like log5 x³ = 6. We removed the negative signs.
Third, use the Definition of Common Logarithms. ( b
Fourth, simplify 5
Fifth, take the cube root of both sides. Your problem should look like x = 3√15625.
Sixth, calculate your problem to get x = 25 which is the answer.
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Answer:
1. value is 0; x-3 is a factor . . . . . . . . . . . . . .third choice
2. evaluates at x = -1; remainder is -11 . . . . first choice
Step-by-step explanation:
Dividing f(x) by (x -a) gives ...
f(x)/(x -a) = g(x) +r/(x -a) . . . . some quotient and a remainder r
If we multiply this expression by (x -a), we see ...
f(x) = (x -a)g(x) +r
so
f(a) = (a -a)g(a) +r . . . . . evaluate the above equation at x=a
f(a) = 0 +r
f(a) = r . . . . . . . . . a statement of the remainder theorem
If r=0, then x-a is a factor of f(x) = (x-a)g(x).
___
1. We have "a" = 3, and f(3) = 0. Therefore (x-3) is a factor.
__
2. We have "a" = -1, and f(-1) = -11. Therefore the remainder from division by (x+1) is -11.
Answer:
Proof in explanation.
Step-by-step explanation:
I'm going to attempt this by squeeze theorem.
We know that is a variable number between -1 and 1 (inclusive).
This means that .
for all value
. So if we multiply all sides of our inequality by this, it will not effect the direction of the inequalities.
By squeeze theorem, if
and , then we can also conclude that
.
So we can actually evaluate the "if" limits pretty easily since both are continuous and exist at .
.
We can finally conclude that by squeeze theorem.
Some people call this sandwich theorem.