Question Show that for positive real numbers $u$ and $v$ , and a positive real number $a \neq 1$ , $\log_{a}\left(\frac{u}{v}\right) = \log_{a}u - \log_{a}v$ .

Studdy Solution

STEP 1

Assumptions
1. $a$ is a positive real number and $a \neq 1$ .
2. $u$ and $v$ are positive real numbers.
3. We are using the properties of logarithms to prove the given statement.

STEP 2

Recall the quotient rule for logarithms, which states that the logarithm of a quotient is equal to the difference of the logarithms of the numerator and the denominator, with the same base.
$\log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y$

STEP 3

We will use the quotient rule to show that $\log _{a}\left(\frac{u}{v}\right)$ is equal to $\log _{a} u - \log _{a} v$ .

STEP 4

Let's express $\log _{a} u$ and $\log _{a} v$ in exponential form to understand their relationship with $\frac{u}{v}$ .
$a^{\log _{a} u} = u$ $a^{\log _{a} v} = v$

STEP 5

Now, we will use the properties of exponents to divide $u$ by $v$ .
$\frac{u}{v} = \frac{a^{\log _{a} u}}{a^{\log _{a} v}}$

STEP 6

Apply the property of exponents that states $\frac{a^m}{a^n} = a^{m-n}$ .
$\frac{u}{v} = a^{\log _{a} u - \log _{a} v}$

STEP 7

Take the logarithm with base $a$ of both sides of the equation.
$\log _{a} \left(\frac{u}{v}\right) = \log _{a} \left(a^{\log _{a} u - \log _{a} v}\right)$

STEP 8

Apply the property of logarithms that states $\log_b (b^x) = x$ .
$\log _{a} \left(\frac{u}{v}\right) = \log _{a} u - \log _{a} v$

STEP 9

We have shown that $\log _{a}\left(\frac{u}{v}\right)$ is indeed equal to $\log _{a} u - \log _{a} v$ , using the properties of logarithms and exponents.
This completes the proof.

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Question Show that for positive real numbers $u$ and $v$ , and a positive real number $a \neq 1$ , $\log_{a}\left(\frac{u}{v}\right) = \log_{a}u - \log_{a}v$ .

Studdy Solution

STEP 1

Assumptions
1. $a$ is a positive real number and $a \neq 1$ .
2. $u$ and $v$ are positive real numbers.
3. We are using the properties of logarithms to prove the given statement.

STEP 2

Recall the quotient rule for logarithms, which states that the logarithm of a quotient is equal to the difference of the logarithms of the numerator and the denominator, with the same base.
$\log_b \left(\frac{x}{y}\right) = \log_b x - \log_b y$

STEP 3

We will use the quotient rule to show that $\log _{a}\left(\frac{u}{v}\right)$ is equal to $\log _{a} u - \log _{a} v$ .

STEP 4

Let's express $\log _{a} u$ and $\log _{a} v$ in exponential form to understand their relationship with $\frac{u}{v}$ .
$a^{\log _{a} u} = u$ $a^{\log _{a} v} = v$

STEP 5

Now, we will use the properties of exponents to divide $u$ by $v$ .
$\frac{u}{v} = \frac{a^{\log _{a} u}}{a^{\log _{a} v}}$

STEP 6

Apply the property of exponents that states $\frac{a^m}{a^n} = a^{m-n}$ .
$\frac{u}{v} = a^{\log _{a} u - \log _{a} v}$

STEP 7

Take the logarithm with base $a$ of both sides of the equation.
$\log _{a} \left(\frac{u}{v}\right) = \log _{a} \left(a^{\log _{a} u - \log _{a} v}\right)$

STEP 8

Apply the property of logarithms that states $\log_b (b^x) = x$ .
$\log _{a} \left(\frac{u}{v}\right) = \log _{a} u - \log _{a} v$

STEP 9

We have shown that $\log _{a}\left(\frac{u}{v}\right)$ is indeed equal to $\log _{a} u - \log _{a} v$ , using the properties of logarithms and exponents.
This completes the proof.

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Question Show that for positive real numbers uuu and vvv, and a positive real number a≠1a \neq 1a=1, log⁡a(uv)=log⁡au−log⁡av\log_{a}\left(\frac{u}{v}\right) = \log_{a}u - \log_{a}vloga​(vu​)=loga​u−loga​v.

Studdy Solution

STEP 1

Assumptions1. aaa is a positive real number and a≠1a \neq 1a=1.2. uuu and vvv are positive real numbers.3. We are using the properties of logarithms to prove the given statement.

STEP 2

STEP 3

We will use the quotient rule to show that log⁡a(uv)\log _{a}\left(\frac{u}{v}\right)loga​(vu​) is equal to log⁡au−log⁡av\log _{a} u - \log _{a} vloga​u−loga​v.

STEP 4

Let's express log⁡au\log _{a} uloga​u and log⁡av\log _{a} vloga​v in exponential form to understand their relationship with uv\frac{u}{v}vu​.alog⁡au=ua^{\log _{a} u} = ualoga​u=u alog⁡av=va^{\log _{a} v} = valoga​v=v

STEP 5

Now, we will use the properties of exponents to divide uuu by vvv.uv=alog⁡aualog⁡av\frac{u}{v} = \frac{a^{\log _{a} u}}{a^{\log _{a} v}}vu​=aloga​valoga​u​

STEP 6

Apply the property of exponents that states aman=am−n\frac{a^m}{a^n} = a^{m-n}anam​=am−n.uv=alog⁡au−log⁡av\frac{u}{v} = a^{\log _{a} u - \log _{a} v}vu​=aloga​u−loga​v

STEP 7

Take the logarithm with base aaa of both sides of the equation.log⁡a(uv)=log⁡a(alog⁡au−log⁡av)\log _{a} \left(\frac{u}{v}\right) = \log _{a} \left(a^{\log _{a} u - \log _{a} v}\right)loga​(vu​)=loga​(aloga​u−loga​v)

STEP 8

Apply the property of logarithms that states log⁡b(bx)=x\log_b (b^x) = xlogb​(bx)=x.log⁡a(uv)=log⁡au−log⁡av\log _{a} \left(\frac{u}{v}\right) = \log _{a} u - \log _{a} vloga​(vu​)=loga​u−loga​v

STEP 9

We have shown that log⁡a(uv)\log _{a}\left(\frac{u}{v}\right)loga​(vu​) is indeed equal to log⁡au−log⁡av\log _{a} u - \log _{a} vloga​u−loga​v, using the properties of logarithms and exponents.This completes the proof.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Question Show that for positive real numbers $u$ and $v$ , and a positive real number $a \neq 1$ , $\log_{a}\left(\frac{u}{v}\right) = \log_{a}u - \log_{a}v$ .

Assumptions
1. $a$ is a positive real number and $a \neq 1$ .
2. $u$ and $v$ are positive real numbers.
3. We are using the properties of logarithms to prove the given statement.

We will use the quotient rule to show that $\log _{a}\left(\frac{u}{v}\right)$ is equal to $\log _{a} u - \log _{a} v$ .

Let's express $\log _{a} u$ and $\log _{a} v$ in exponential form to understand their relationship with $\frac{u}{v}$ .
$a^{\log _{a} u} = u$ $a^{\log _{a} v} = v$

Now, we will use the properties of exponents to divide $u$ by $v$ .
$\frac{u}{v} = \frac{a^{\log _{a} u}}{a^{\log _{a} v}}$

Apply the property of exponents that states $\frac{a^m}{a^n} = a^{m-n}$ .
$\frac{u}{v} = a^{\log _{a} u - \log _{a} v}$

Take the logarithm with base $a$ of both sides of the equation.
$\log _{a} \left(\frac{u}{v}\right) = \log _{a} \left(a^{\log _{a} u - \log _{a} v}\right)$

Apply the property of logarithms that states $\log_b (b^x) = x$ .
$\log _{a} \left(\frac{u}{v}\right) = \log _{a} u - \log _{a} v$

We have shown that $\log _{a}\left(\frac{u}{v}\right)$ is indeed equal to $\log _{a} u - \log _{a} v$ , using the properties of logarithms and exponents.
This completes the proof.