MATH SOLVE

10 months ago

Q:
# Which system of equation can represent the equation?

Accepted Solution

A:

Answer:

log(x+3)

y₁ = --------------

log(4)

log(2 + x)

y₂ = ----------------

log(2)

Explanation:

1) Here you must use the property named change of base of the logarithm.

That rule is:

[tex]log_{a}b= \frac{log_c{b}}{log_ca} [/tex]

2) Changing to base 10 that is:

[tex]log_{b}a= \frac{loga}{logb} [/tex]

3) Using it with the given equation, you get:

[tex]log_4(x+3)= \frac{log_(x+3)}{log(4)} log_2(2+x)= \frac{log(2+x)}{log(2)}[/tex]

4) Now, to set your system you can make the left side of the equation equal to y₁ and the right side equal to y₂

log(x+3)

y₁ = -------------

log(4)

log(2 + x)

y₂ = ---------------

log(2)

And that is your system which is the first choice of the list.

log(x+3)

y₁ = --------------

log(4)

log(2 + x)

y₂ = ----------------

log(2)

Explanation:

1) Here you must use the property named change of base of the logarithm.

That rule is:

[tex]log_{a}b= \frac{log_c{b}}{log_ca} [/tex]

2) Changing to base 10 that is:

[tex]log_{b}a= \frac{loga}{logb} [/tex]

3) Using it with the given equation, you get:

[tex]log_4(x+3)= \frac{log_(x+3)}{log(4)} log_2(2+x)= \frac{log(2+x)}{log(2)}[/tex]

4) Now, to set your system you can make the left side of the equation equal to y₁ and the right side equal to y₂

log(x+3)

y₁ = -------------

log(4)

log(2 + x)

y₂ = ---------------

log(2)

And that is your system which is the first choice of the list.