Math / Algebra

Question8. Simplify/solve the logarithm (a). $\log _{(1 / 4)} 32$ (b). $\log x-3 \sqrt{\log x}+2=0$ (c). $8^{-x+11}=16^{2 x}$ (d). Write as sum and difference of logarithm $\ln \left[\frac{x^{2}-x-2}{(x+4)^{2}}\right]^{1 / 3}$ for $x>2$ (e). Write as single $\operatorname{logarithm} 3 \log _{5}(3 x+1)-2 \log _{5}(2 x-1)-\log _{5} x$

Studdy Solution

STEP 1

What is this asking? We're tackling some exciting logarithm problems, simplifying, solving, and rewriting them in different forms! Watch out! Remember those logarithm rules and be careful with those bases and exponents!

STEP 2

1. Solve $\log _{(1 / 4)} 32$
2. Solve $\log x-3 \sqrt{\log x}+2=0$
3. Solve $8^{-x+11}=16^{2 x}$
4. Expand $\ln \left[\frac{x^{2}-x-2}{(x+4)^{2}}\right]^{1 / 3}$
5. Condense $3 \log _{5}(3 x+1)-2 \log _{5}(2 x-1)-\log _{5} x$

STEP 3

Let's set our logarithm equal to $y$ , so $\log _{(1 / 4)} 32 = y$ .
This means $(1/4)^y = 32$ .
We're doing this to make the equation easier to work with!

STEP 4

Now, let's rewrite everything in terms of base 2.
We know $1/4 = 2^{-2}$ and $32 = 2^5$ .
So, we have $(2^{-2})^y = 2^5$ .

STEP 5

Using the power rule of exponents, we get $2^{-2y} = 2^5$ .
Since the bases are the same, we can equate the exponents: $-2y = 5$ .

STEP 6

Finally, divide both sides by $-2$ to solve for $y$ : $y = -\frac{5}{2}$ .
So, $\log _{(1 / 4)} 32 = -\frac{5}{2}$ .

STEP 7

Let $u = \sqrt{\log x}$ .
This substitution makes our equation look much simpler: $u^2 - 3u + 2 = 0$ .

STEP 8

Now, we can factor this quadratic equation: $(u-1)(u-2) = 0$ .

STEP 9

This gives us two possible solutions for $u$ : $u=1$ or $u=2$ .

STEP 10

Let's substitute back $\sqrt{\log x}$ for $u$ .
If $\sqrt{\log x} = 1$ , then $\log x = 1$ , which means $x = 10$ .
If $\sqrt{\log x} = 2$ , then $\log x = 4$ , which means $x = 10^4 = 10000$ .
We're assuming here that $\log$ means $\log_{10}$ .

STEP 11

Let's rewrite both sides of the equation with the same base.
We know $8 = 2^3$ and $16 = 2^4$ , so we have $(2^3)^{-x+11} = (2^4)^{2x}$ .

STEP 12

Using the power rule, we get $2^{-3x+33} = 2^{8x}$ .

STEP 13

Since the bases are equal, we can equate the exponents: $-3x + 33 = 8x$ .

STEP 14

Now, add $3x$ to both sides: $33 = 11x$ .

STEP 15

Finally, divide both sides by 11 to find $x = 3$ .

STEP 16

Using the power rule of logarithms, we can bring the exponent $\frac{1}{3}$ down in front: $\frac{1}{3} \ln \left[\frac{x^{2}-x-2}{(x+4)^{2}}\right]$ .

STEP 17

Next, we can use the quotient rule to expand the logarithm: $\frac{1}{3} [\ln(x^2 - x - 2) - \ln(x+4)^2]$ .

STEP 18

We can factor the quadratic expression: $x^2 - x - 2 = (x-2)(x+1)$ .
So, we have $\frac{1}{3} [\ln((x-2)(x+1)) - \ln(x+4)^2]$ .

STEP 19

Using the product rule and power rule, we can expand further: $\frac{1}{3} [\ln(x-2) + \ln(x+1) - 2\ln(x+4)]$ .

STEP 20

Finally, we distribute the $\frac{1}{3}$ : $\frac{1}{3}\ln(x-2) + \frac{1}{3}\ln(x+1) - \frac{2}{3}\ln(x+4)$ .

STEP 21

Using the power rule, we can rewrite the coefficients as exponents: $\log _{5}(3 x+1)^3 - \log _{5}(2 x-1)^2 - \log _{5} x$ .

STEP 22

Now, we can use the quotient rule to combine the first two logarithms: $\log _{5} \frac{(3 x+1)^3}{(2 x-1)^2} - \log _{5} x$ .

STEP 23

Finally, we use the quotient rule again to combine the remaining logarithms into a single logarithm: $\log _{5} \frac{(3 x+1)^3}{x(2 x-1)^2}$ .

STEP 24

(a) $-\frac{5}{2}$ (b) $x = 10$ and $x = 10000$ (c) $x = 3$ (d) $\frac{1}{3}\ln(x-2) + \frac{1}{3}\ln(x+1) - \frac{2}{3}\ln(x+4)$ (e) $\log _{5} \frac{(3 x+1)^3}{x(2 x-1)^2}$

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Studdy Solution

STEP 1

What is this asking? We're tackling some exciting logarithm problems, simplifying, solving, and rewriting them in different forms! Watch out! Remember those logarithm rules and be careful with those bases and exponents!

STEP 2

STEP 3

Let's **set** our logarithm equal to yyy, so log⁡(1/4)32=y\log _{(1 / 4)} 32 = ylog(1/4)​32=y.This means (1/4)y=32(1/4)^y = 32(1/4)y=32.We're doing this to make the equation easier to work with!

STEP 4

Now, let's **rewrite** everything in terms of base 2.We know 1/4=2−21/4 = 2^{-2}1/4=2−2 and 32=2532 = 2^532=25.So, we have (2−2)y=25(2^{-2})^y = 2^5(2−2)y=25.

STEP 5

Using the power rule of exponents, we get 2−2y=252^{-2y} = 2^52−2y=25.Since the bases are the same, we can **equate** the exponents: −2y=5-2y = 5−2y=5.

STEP 6

Finally, **divide** both sides by −2-2−2 to **solve** for yyy: y=−52y = -\frac{5}{2}y=−25​.So, log⁡(1/4)32=−52\log _{(1 / 4)} 32 = -\frac{5}{2}log(1/4)​32=−25​.

STEP 7

Let u=log⁡xu = \sqrt{\log x}u=logx​.This **substitution** makes our equation look much simpler: u2−3u+2=0u^2 - 3u + 2 = 0u2−3u+2=0.

STEP 8

Now, we can **factor** this quadratic equation: (u−1)(u−2)=0(u-1)(u-2) = 0(u−1)(u−2)=0.

STEP 9

This gives us two possible solutions for uuu: u=1u=1u=1 or u=2u=2u=2.

STEP 10

STEP 11

Let's **rewrite** both sides of the equation with the same base.We know 8=238 = 2^38=23 and 16=2416 = 2^416=24, so we have (23)−x+11=(24)2x(2^3)^{-x+11} = (2^4)^{2x}(23)−x+11=(24)2x.

STEP 12

Using the power rule, we get 2−3x+33=28x2^{-3x+33} = 2^{8x}2−3x+33=28x.

STEP 13

Since the bases are equal, we can **equate** the exponents: −3x+33=8x-3x + 33 = 8x−3x+33=8x.

STEP 14

Now, **add** 3x3x3x to both sides: 33=11x33 = 11x33=11x.

STEP 15

Finally, **divide** both sides by 11 to find x=3x = 3x=3.

STEP 16

Using the power rule of logarithms, we can **bring** the exponent 13\frac{1}{3}31​ down in front: 13ln⁡[x2−x−2(x+4)2]\frac{1}{3} \ln \left[\frac{x^{2}-x-2}{(x+4)^{2}}\right]31​ln[(x+4)2x2−x−2​].

STEP 17

Next, we can use the quotient rule to **expand** the logarithm: 13[ln⁡(x2−x−2)−ln⁡(x+4)2]\frac{1}{3} [\ln(x^2 - x - 2) - \ln(x+4)^2]31​[ln(x2−x−2)−ln(x+4)2].

STEP 18

We can **factor** the quadratic expression: x2−x−2=(x−2)(x+1)x^2 - x - 2 = (x-2)(x+1)x2−x−2=(x−2)(x+1).So, we have 13[ln⁡((x−2)(x+1))−ln⁡(x+4)2]\frac{1}{3} [\ln((x-2)(x+1)) - \ln(x+4)^2]31​[ln((x−2)(x+1))−ln(x+4)2].

STEP 19

Using the product rule and power rule, we can **expand** further: 13[ln⁡(x−2)+ln⁡(x+1)−2ln⁡(x+4)]\frac{1}{3} [\ln(x-2) + \ln(x+1) - 2\ln(x+4)]31​[ln(x−2)+ln(x+1)−2ln(x+4)].

STEP 20

Finally, we **distribute** the 13\frac{1}{3}31​: 13ln⁡(x−2)+13ln⁡(x+1)−23ln⁡(x+4)\frac{1}{3}\ln(x-2) + \frac{1}{3}\ln(x+1) - \frac{2}{3}\ln(x+4)31​ln(x−2)+31​ln(x+1)−32​ln(x+4).

STEP 21

Using the power rule, we can **rewrite** the coefficients as exponents: log⁡5(3x+1)3−log⁡5(2x−1)2−log⁡5x\log _{5}(3 x+1)^3 - \log _{5}(2 x-1)^2 - \log _{5} xlog5​(3x+1)3−log5​(2x−1)2−log5​x.

STEP 22

Now, we can use the quotient rule to **combine** the first two logarithms: log⁡5(3x+1)3(2x−1)2−log⁡5x\log _{5} \frac{(3 x+1)^3}{(2 x-1)^2} - \log _{5} xlog5​(2x−1)2(3x+1)3​−log5​x.

STEP 23

Finally, we use the quotient rule again to **combine** the remaining logarithms into a single logarithm: log⁡5(3x+1)3x(2x−1)2\log _{5} \frac{(3 x+1)^3}{x(2 x-1)^2}log5​x(2x−1)2(3x+1)3​.

STEP 24

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Let's set our logarithm equal to $y$ , so $\log _{(1 / 4)} 32 = y$ .
This means $(1/4)^y = 32$ .
We're doing this to make the equation easier to work with!

Now, let's rewrite everything in terms of base 2.
We know $1/4 = 2^{-2}$ and $32 = 2^5$ .
So, we have $(2^{-2})^y = 2^5$ .

Using the power rule of exponents, we get $2^{-2y} = 2^5$ .
Since the bases are the same, we can equate the exponents: $-2y = 5$ .

Finally, divide both sides by $-2$ to solve for $y$ : $y = -\frac{5}{2}$ .
So, $\log _{(1 / 4)} 32 = -\frac{5}{2}$ .

Let $u = \sqrt{\log x}$ .
This substitution makes our equation look much simpler: $u^2 - 3u + 2 = 0$ .

Now, we can factor this quadratic equation: $(u-1)(u-2) = 0$ .

This gives us two possible solutions for $u$ : $u=1$ or $u=2$ .

Let's rewrite both sides of the equation with the same base.
We know $8 = 2^3$ and $16 = 2^4$ , so we have $(2^3)^{-x+11} = (2^4)^{2x}$ .

Using the power rule, we get $2^{-3x+33} = 2^{8x}$ .

Since the bases are equal, we can equate the exponents: $-3x + 33 = 8x$ .

Now, add $3x$ to both sides: $33 = 11x$ .

Finally, divide both sides by 11 to find $x = 3$ .

Using the power rule of logarithms, we can bring the exponent $\frac{1}{3}$ down in front: $\frac{1}{3} \ln \left[\frac{x^{2}-x-2}{(x+4)^{2}}\right]$ .

Next, we can use the quotient rule to expand the logarithm: $\frac{1}{3} [\ln(x^2 - x - 2) - \ln(x+4)^2]$ .

We can factor the quadratic expression: $x^2 - x - 2 = (x-2)(x+1)$ .
So, we have $\frac{1}{3} [\ln((x-2)(x+1)) - \ln(x+4)^2]$ .

Using the product rule and power rule, we can expand further: $\frac{1}{3} [\ln(x-2) + \ln(x+1) - 2\ln(x+4)]$ .

Finally, we distribute the $\frac{1}{3}$ : $\frac{1}{3}\ln(x-2) + \frac{1}{3}\ln(x+1) - \frac{2}{3}\ln(x+4)$ .

Using the power rule, we can rewrite the coefficients as exponents: $\log _{5}(3 x+1)^3 - \log _{5}(2 x-1)^2 - \log _{5} x$ .

Now, we can use the quotient rule to combine the first two logarithms: $\log _{5} \frac{(3 x+1)^3}{(2 x-1)^2} - \log _{5} x$ .

Finally, we use the quotient rule again to combine the remaining logarithms into a single logarithm: $\log _{5} \frac{(3 x+1)^3}{x(2 x-1)^2}$ .