Use the law of exponents to rewrite the following as a sum or difference and without exponents: a) log base a of (x^2/yz^3) b) log (x/cube root of 1-x)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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I need to rewrite the answer b) such that:

`log (x/root(3)(1-x)) = log x - (1/3)log(1-x)`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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a) You need to use the quotient law to convert the logarithm of quotient in a difference of two logarithms such that:

`log_a (x^2/(yz^3)) = log_a x^2 - log_a (yz^3)`

You need to convert the logarithm of product in a sum of two logarithms such that:

`log_a (yz^3) = log_a y + log_a (z^3)`

You need to use exponent law to write `log_a x^2`  and `log_a (z^3) ` such that:

`log_a (x^2/(yz^3)) = 2log_a x -log_a y - 3log_a z`

Hence, evaluating `log_a (x^2/(yz^3))`  yields `log_a (x^2/(yz^3)) = 2log_a x - log_a y - 3log_a z` .

b) You need to use the quotient law to convert the logarithm of quotient in a difference of two logarithms such that:

`log x/(root(3)(1-x)) = log x - log (root(3)(1-x))`

You need to remember that `(root(3)(1-x)) = (1-x)^(1/3)`

`log x/(root(3)(1-x)) = log x - log (1-x)^(1/3)`

`log x/(root(3)(1-x)) = log x - (1/3)*log (1-x)`

Hence, evaluating`log x/(root(3)(1-x)) ` yields `log x/(root(3)(1-x)) = log x - (1/3)*log (1-x).`

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