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Math  /  Algebra

Question4. 32−7x=22x−13^{2-7 x}=2^{2 x-1}
5. 2e2x+13ex−7=02 e^{2 x}+13 e^{x}-7=0
6. log⁡8(x+6)−log⁡8x=log⁡858\log _{8}(x+6)-\log _{8} x=\log _{8} 58
7. ln⁡x+ln⁡(x−3)=ln⁡(7x)\ln x+\ln (x-3)=\ln (7 x)

Studdy Solution

STEP 1

What is this asking? We've got a bunch of equations with exponents and logarithms, and we need to find the values of xx that make them true! Watch out! Remember that logarithms have domain restrictions!
We can't take the log of a negative number or zero.
Keep an eye out for those sneaky extraneous solutions!

STEP 2

1. Exponential Equation with Different Bases
2. Exponential Equation in Quadratic Form
3. Logarithmic Equation
4. Logarithmic Equation with Extraneous Solution

STEP 3

Alright, let's tackle 32−7x=22x−13^{2-7x} = 2^{2x-1}!
To get those exponents down, we'll **take the natural logarithm (ln) of both sides**: ln⁡(32−7x)=ln⁡(22x−1) \ln(3^{2-7x}) = \ln(2^{2x-1}) Remember, we can do this because applying the same function to both sides of an equation maintains equality!

STEP 4

Now, let's use that super handy **power rule of logarithms**: ln⁡(ab)=b⋅ln⁡(a)\ln(a^b) = b \cdot \ln(a).
This lets us bring those exponents down in front: (2−7x)⋅ln⁡(3)=(2x−1)⋅ln⁡(2) (2-7x) \cdot \ln(3) = (2x-1) \cdot \ln(2) Much better!

STEP 5

**Distribute** ln⁡(3)\ln(3) and ln⁡(2)\ln(2) on both sides: 2⋅ln⁡(3)−7x⋅ln⁡(3)=2x⋅ln⁡(2)−ln⁡(2) 2 \cdot \ln(3) - 7x \cdot \ln(3) = 2x \cdot \ln(2) - \ln(2)

STEP 6

Let's **gather all the terms with** xx **on one side** and all the **constant terms on the other**: 2⋅ln⁡(3)+ln⁡(2)=2x⋅ln⁡(2)+7x⋅ln⁡(3) 2 \cdot \ln(3) + \ln(2) = 2x \cdot \ln(2) + 7x \cdot \ln(3)

STEP 7

**Factor out** xx on the right side: 2⋅ln⁡(3)+ln⁡(2)=x(2⋅ln⁡(2)+7⋅ln⁡(3)) 2 \cdot \ln(3) + \ln(2) = x(2 \cdot \ln(2) + 7 \cdot \ln(3))

STEP 8

Finally, **divide both sides** by (2⋅ln⁡(2)+7⋅ln⁡(3))(2 \cdot \ln(2) + 7 \cdot \ln(3)) to **isolate** xx: x=2⋅ln⁡(3)+ln⁡(2)2⋅ln⁡(2)+7⋅ln⁡(3) x = \frac{2 \cdot \ln(3) + \ln(2)}{2 \cdot \ln(2) + 7 \cdot \ln(3)}

STEP 9

Now, we can plug this into a calculator to get an approximate value for xx: x≈2⋅1.0986+0.69312⋅0.6931+7⋅1.0986≈2.89038.0834≈0.3575 x \approx \frac{2 \cdot 1.0986 + 0.6931}{2 \cdot 0.6931 + 7 \cdot 1.0986} \approx \frac{2.8903}{8.0834} \approx 0.3575 So, x≈0.3575x \approx \mathbf{0.3575}!

STEP 10

Let's look at 2e2x+13ex−7=02e^{2x} + 13e^x - 7 = 0.
Notice how this looks like a quadratic equation?
Let's make a **substitution**: Let u=exu = e^x.
Then u2=(ex)2=e2xu^2 = (e^x)^2 = e^{2x}.
Our equation becomes: 2u2+13u−7=0 2u^2 + 13u - 7 = 0

STEP 11

Now we can **factor this quadratic**: (2u−1)(u+7)=0 (2u - 1)(u + 7) = 0

STEP 12

This gives us two possible solutions for uu: 2u−1=0⇒u=12 2u - 1 = 0 \Rightarrow u = \frac{1}{2} u+7=0⇒u=−7 u + 7 = 0 \Rightarrow u = -7

STEP 13

Remember, u=exu = e^x.
Since exe^x is always positive, u=−7u = -7 is not a valid solution.
So, we only have: ex=12 e^x = \frac{1}{2}

STEP 14

**Take the natural logarithm of both sides**: ln⁡(ex)=ln⁡(12) \ln(e^x) = \ln\left(\frac{1}{2}\right)

STEP 15

Using the property ln⁡(ex)=x\ln(e^x) = x, we get: x=ln⁡(12) x = \ln\left(\frac{1}{2}\right)

STEP 16

We can rewrite this as: x=ln⁡(1)−ln⁡(2) x = \ln(1) - \ln(2) Since ln⁡(1)=0\ln(1) = 0, we have: x=−ln⁡(2)≈−0.6931 x = -\ln(2) \approx \mathbf{-0.6931}

STEP 17

For log⁡8(x+6)−log⁡8(x)=log⁡8(58)\log_8(x+6) - \log_8(x) = \log_8(58), we can use the **quotient rule of logarithms**: log⁡b(a)−log⁡b(c)=log⁡b(ac)\log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right): log⁡8(x+6x)=log⁡8(58) \log_8\left(\frac{x+6}{x}\right) = \log_8(58)

STEP 18

Since the bases are the same, we can **set the arguments equal to each other**: x+6x=58 \frac{x+6}{x} = 58

STEP 19

**Multiply both sides by** xx: x+6=58x x + 6 = 58x

STEP 20

**Subtract** xx **from both sides**: 6=57x 6 = 57x

STEP 21

**Divide both sides by** 5757: x=657=219 x = \frac{6}{57} = \frac{2}{19} So, x=219x = \mathbf{\frac{2}{19}}!

STEP 22

Let's solve ln⁡(x)+ln⁡(x−3)=ln⁡(7x)\ln(x) + \ln(x-3) = \ln(7x).
Using the **product rule of logarithms**: ln⁡(a)+ln⁡(b)=ln⁡(ab)\ln(a) + \ln(b) = \ln(ab): ln⁡(x(x−3))=ln⁡(7x) \ln(x(x-3)) = \ln(7x)

STEP 23

Since the bases are the same, we can **set the arguments equal**: x(x−3)=7x x(x-3) = 7x

STEP 24

**Expand the left side**: x2−3x=7x x^2 - 3x = 7x

STEP 25

**Subtract** 7x7x **from both sides**: x2−10x=0 x^2 - 10x = 0

STEP 26

**Factor out** xx: x(x−10)=0 x(x - 10) = 0

STEP 27

This gives us two possible solutions: x=0x = 0 and x=10x = 10.
However, if we plug x=0x = 0 back into the original equation, we get ln⁡(0)\ln(0), which is undefined!
So, x=0x = 0 is an **extraneous solution**.

STEP 28

Therefore, the only valid solution is x=10x = \mathbf{10}!

STEP 29

Here are our solutions: x≈0.3575x \approx 0.3575 x≈−0.6931x \approx -0.6931x=219x = \frac{2}{19}x=10x = 10

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