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3 years ago

12

Write the following equation in logarithmic form a^(4b-6)=3c+d

1 answer:

Tomtit [17]3 years ago

3 0

Hello from MrBillDoesMath!

Answer: (4b-6) log(a) = log (3c + d)

Discussion:

Take the log of both sides of the equation:

log ( a ^(4b-6)) = log (3c + d)

As the log functions causes exponents to become multiplexers, this equation is the same as

(4b-6) log(a) = log (3c + d)

Thank you,

MrB

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4 years ago

Given f(x) = 3x - 1 and g(x)= -x + 6, find f(-2) + g(5).<br> A. -6<br> B. 6<br> C. 8

marta [7]

Answer:

A. -6

Step-by-step explanation:

$f(x)=3x-1\\\\$

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$f(-2)=-7$

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3 years ago

Evaluate the function at the indicated values if possible. If an indicated value is not in the domain, say so.

goldfiish [28.3K]

Answer:

f(-7)=0.

f(2)=-9/5.

f(3) doesn't exist because 3 isn't in the domain of the function.

Step-by-step explanation:

$f(x)=\frac{x+7}{x^2-9}$ is the given function.

We are asked to find:

$f(-7)$

$f(2)$

$f(3)$ .

f(-7) means to replace x in the expression called f with -7:

Evaluate $\frac{x+7}{x^2-9}$ at $x=-7$

$\frac{(-7)+7}{(-7)^2-9}$

$\frac{0}{49-9}$

$\frac{0}{40}$

$0$

So f(-7)=0.

f(2) means to replace x in the expression called f with 2:

Evaluate $\frac{x+7}{x^2-9}$ at $x=2$

$\frac{2+7}{2^2-9}$

$\frac{9}{4-9}$

$\frac{9}{-5}$

$\frac{-9}{5}$

So f(2)=-9/5

f(3) means to replace x in the expression called f with 3:

Evaluate $\frac{x+7}{x^2-9}$ at $x=3$

$\frac{3+7}{3^2-9}$

$\frac{10}{9-9}$

$\frac{10}{0}$

Division by 0 is not allowed so 3 is not in the domain of our function.

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