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Answer:
D. y 1 = StartFraction log x Over log 6 EndFraction, y 2 = StartFraction log (x + 4) Over log 2 EndFraction
Step-by-step explanation:
Omar should use the system of equations y₁ = logx/log6, and y₂ = log(x+4)/log2 option second is correct.
What is a logarithm?
It is another way to represent the power of numbers, and we say that 'b' is the logarithm of 'c' with base 'a' if and only if 'a' to the power 'b' equals 'c'.
[tex]\rm a^b = c\\log_ac =b[/tex]
We have a log equation:
log₆x = log₂(x + 4)
As we know:
logₐᵇ = log(b)/log(a)
After applying the above property:
logx/log6 = log(x+4)/log2
y₁ = logx/log6
y₂ = log(x+4)/log2
Thus, Omar should use the system of equations y₁ = logx/log6, and y₂ = log(x+4)/log2 option second is correct.
Learn more about the Logarithm here:
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