SOLUTION: Arjun incorrectly writes log7 x+log7 y-log7 z as a single logarithm.
log7 x + log7 y - log7 z = log7 (x+y=z)
Where did Arjun make errors? Explain his errors and the properties of
Algebra ->
Exponential-and-logarithmic-functions
-> SOLUTION: Arjun incorrectly writes log7 x+log7 y-log7 z as a single logarithm.
log7 x + log7 y - log7 z = log7 (x+y=z)
Where did Arjun make errors? Explain his errors and the properties of
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Question 1189571: Arjun incorrectly writes log7 x+log7 y-log7 z as a single logarithm.
log7 x + log7 y - log7 z = log7 (x+y=z)
Where did Arjun make errors? Explain his errors and the properties of
logarithms that lead to the correct answer. State the correct answer.
please help, i beg you, please Found 2 solutions by math_tutor2020, Edwin McCravy:Answer by math_tutor2020(3817) (Show Source):
If so, then that equation is incorrect (or Arjun is incorrect in writing such an equation).
The log rules we'll use are
log(A) + log(B) = log(A*B)
log(A) - log(B) = log(A/B)
These rules are used frequently, so be sure to memorize them or have them handy on a flashcard somewhere.
Based on the first rule, we can say
log7 x + log7 y = log7 (x*y)
Then incoporating the second rule allows us to say
log7 (x*y) - log7 z = log7 (x*y/z)
In short, the original expression of
log7 x + log7 y - log7 z
becomes
log7 (x*y/z)
Side note: By writing "log7", it means "log base 7".
You can put this solution on YOUR website!
Arjun incorrectly writes log7 x+log7 y-log7 z as a single logarithm.
log7 x + log7 y - log7 z = log7 (x+y=z)
Where did Arjun make errors? Explain his errors and the properties of
logarithms that leads to the correct answer. State the correct answer.
The correct answer is
The properties of logarithms to use are
Arjun just doesn't understand the properties of logarithms.
Edwin
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