Math  /  Algebra

QuestionExpand the logarithm fully using the properties of logs. Express the final answer in terms of logx\log{x}, logy\log{y}, and logz\log{z}.
logyzx53\log{\frac{y}{\sqrt[3]{zx^5}}}

Studdy Solution

STEP 1

What is this asking? We need to rewrite a *single* logarithm as the *sum or difference* of several simpler logarithms. Watch out! Remember the rules of logarithms, especially how the cube root and exponents play a role!

STEP 2

1. Rewrite the cube root as an exponent
2. Expand the quotient inside the logarithm
3. Expand the product inside the logarithm
4. Expand the power inside the logarithm

STEP 3

We have a cube root in the denominator, which we can rewrite as an exponent of 13\frac{1}{3}.
This makes it easier to apply the logarithm power rule later.
So, zx53\sqrt[3]{zx^5} becomes (zx5)13(zx^5)^{\frac{1}{3}}.
This gives us the new expression:
logy(zx5)13 \log{\frac{y}{(zx^5)^{\frac{1}{3}}}}

STEP 4

Remember the quotient rule: log(ab)=logalogb\log(\frac{a}{b}) = \log{a} - \log{b}.
Let's apply this to our expression, where a=ya = y and b=(zx5)13b = (zx^5)^{\frac{1}{3}}:
logylog(zx5)13 \log{y} - \log{(zx^5)^{\frac{1}{3}}}

STEP 5

We have a product inside the second logarithm term.
Remember the product rule: log(ab)=loga+logb\log(a \cdot b) = \log{a} + \log{b}.
Here, a=za = z and b=x5b = x^5, so log(zx5)13\log{(zx^5)^{\frac{1}{3}}} becomes log(z13x53)\log{(z^{\frac{1}{3}}x^{\frac{5}{3}})}.

STEP 6

Now apply the product rule:
logy(logz13+logx53) \log{y} - (\log{z^{\frac{1}{3}}} + \log{x^{\frac{5}{3}}})

STEP 7

Distribute the negative sign:
logylogz13logx53 \log{y} - \log{z^{\frac{1}{3}}} - \log{x^{\frac{5}{3}}}

STEP 8

Time for the power rule!
Remember, logab=bloga\log{a^b} = b\log{a}.
Let's apply this to the last two terms.

STEP 9

For the term logz13\log{z^{\frac{1}{3}}}, we have a=za = z and b=13b = \frac{1}{3}, so it becomes 13logz\frac{1}{3}\log{z}.

STEP 10

For the term logx53\log{x^{\frac{5}{3}}}, we have a=xa = x and b=53b = \frac{5}{3}, so it becomes 53logx\frac{5}{3}\log{x}.

STEP 11

Putting it all together, we get:
logy13logz53logx \log{y} - \frac{1}{3}\log{z} - \frac{5}{3}\log{x}

STEP 12

The fully expanded logarithm is:
logy13logz53logx \log{y} - \frac{1}{3}\log{z} - \frac{5}{3}\log{x}

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