Asked by Jess
Express x in terms of a,b and c.
log x = 1/2 (log a + log b - log c)
Please solve and explain how to do this type of problem, thank you!
log x = 1/2 (log a + log b - log c)
Please solve and explain how to do this type of problem, thank you!
Answers
Answered by
Anonymous
duplicate
Answered by
MathMate
To solve
log x = 1/2 (log a + log b - log c)
we need to know some properties of logarithms, namely:
1. log(a)+log(b) = log(ab)
2. (1/2)log(a)=log(a<sup>-1/2</sup>)=log(√(a))
3. e<sup>ln x </sup> = x, or
10<sup>log10 x</sup> = x
Proceeding to simplify the right-hand-side,
log x = 1/2 (log a + log b - log c)
= (1/2)log a + (1/2)log b + (1/2)log c
= log(√a) + log(√b) + log(√c)
= log(√a √b √c)
= log(√(abc))
Assuming log() stands for logarithm to the base e,
e<sup>log x</sup> = e<sup>log(√(abc))</sup>
x = √(abc)
log x = 1/2 (log a + log b - log c)
we need to know some properties of logarithms, namely:
1. log(a)+log(b) = log(ab)
2. (1/2)log(a)=log(a<sup>-1/2</sup>)=log(√(a))
3. e<sup>ln x </sup> = x, or
10<sup>log10 x</sup> = x
Proceeding to simplify the right-hand-side,
log x = 1/2 (log a + log b - log c)
= (1/2)log a + (1/2)log b + (1/2)log c
= log(√a) + log(√b) + log(√c)
= log(√a √b √c)
= log(√(abc))
Assuming log() stands for logarithm to the base e,
e<sup>log x</sup> = e<sup>log(√(abc))</sup>
x = √(abc)
Answered by
MathMate
2nd rule of logarithm should read:
2. (1/2)log(a)=log(a<sup>1/2</sup>)=log(√(a))
and the solution has to be corrected because of an erroneous sign:
Proceeding to simplify the right-hand-side,
log x = 1/2 (log a + log b - log c)
= (1/2)log a + (1/2)log b <b>-</b> (1/2)log c
= log(√a) + log(√b) <b>-</b> log(√c)
= log(√a √b <b>/</b> √c)
= log(√(ab<b>/</b>c))
Assuming log() stands for logarithm to the base e,
e<sup>log x</sup> = e<sup>log(√(ab<b>/</b>c))</sup>
x = √(ab<b>/</b>c)
2. (1/2)log(a)=log(a<sup>1/2</sup>)=log(√(a))
and the solution has to be corrected because of an erroneous sign:
Proceeding to simplify the right-hand-side,
log x = 1/2 (log a + log b - log c)
= (1/2)log a + (1/2)log b <b>-</b> (1/2)log c
= log(√a) + log(√b) <b>-</b> log(√c)
= log(√a √b <b>/</b> √c)
= log(√(ab<b>/</b>c))
Assuming log() stands for logarithm to the base e,
e<sup>log x</sup> = e<sup>log(√(ab<b>/</b>c))</sup>
x = √(ab<b>/</b>c)
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