Math / Algebra

QuestionUse logarithms to solve. Give an exact simplified answer. Enter DNE if there is no solution. $3 \cdot e^{4-5 x}+4=-33$ $x=$ Question Help: Written Example Submit Question

Studdy Solution

STEP 1

What is this asking? We need to find the value of $x$ that makes the equation $3 \cdot e^{4-5x} + 4 = -33$ true, using logarithms, and give the answer in exact simplified form. Watch out! Remember the properties of logarithms and exponents!
Don't get tripped up by the negative sign.

STEP 2

1. Isolate the exponential term
2. Apply the natural logarithm
3. Solve for x

STEP 3

First, we want to get the exponential term all by itself.
Let's subtract $4$ from both sides of the equation: $3 \cdot e^{4-5x} + 4 - 4 = -33 - 4$ $3 \cdot e^{4-5x} = -37$

STEP 4

Next, we divide both sides by $3$ : $\frac{3 \cdot e^{4-5x}}{3} = \frac{-37}{3}$ $e^{4-5x} = -\frac{37}{3}$

STEP 5

Now, we'll apply the natural logarithm (ln) to both sides.
This is the key step that lets us solve for $x$ because logarithms are the inverse of exponential functions: $\ln(e^{4-5x}) = \ln\left(-\frac{37}{3}\right)$

STEP 6

Whoa, hold on!
We have a problem here.
We're trying to take the logarithm of a negative number, $- \frac{37}{3}$ .
This isn't allowed!
The logarithm of a negative number is undefined.

STEP 7

Since we encountered an undefined logarithm, there's no real solution for $x$ that satisfies the original equation.

STEP 8

$x =$ DNE

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QuestionUse logarithms to solve. Give an exact simplified answer. Enter DNE if there is no solution. $3 \cdot e^{4-5 x}+4=-33$ $x=$ Question Help: Written Example Submit Question

Studdy Solution

STEP 1

What is this asking? We need to find the value of $x$ that makes the equation $3 \cdot e^{4-5x} + 4 = -33$ true, using logarithms, and give the answer in exact simplified form. Watch out! Remember the properties of logarithms and exponents!
Don't get tripped up by the negative sign.

STEP 2

1. Isolate the exponential term
2. Apply the natural logarithm
3. Solve for x

STEP 3

First, we want to get the exponential term all by itself.
Let's subtract $4$ from both sides of the equation: $3 \cdot e^{4-5x} + 4 - 4 = -33 - 4$ $3 \cdot e^{4-5x} = -37$

STEP 4

Next, we divide both sides by $3$ : $\frac{3 \cdot e^{4-5x}}{3} = \frac{-37}{3}$ $e^{4-5x} = -\frac{37}{3}$

STEP 5

Now, we'll apply the natural logarithm (ln) to both sides.
This is the key step that lets us solve for $x$ because logarithms are the inverse of exponential functions: $\ln(e^{4-5x}) = \ln\left(-\frac{37}{3}\right)$

STEP 6

Whoa, hold on!
We have a problem here.
We're trying to take the logarithm of a negative number, $- \frac{37}{3}$ .
This isn't allowed!
The logarithm of a negative number is undefined.

STEP 7

Since we encountered an undefined logarithm, there's no real solution for $x$ that satisfies the original equation.

STEP 8

$x =$ DNE

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QuestionUse logarithms to solve. Give an exact simplified answer. Enter DNE if there is no solution. 3⋅e4−5x+4=−333 \cdot e^{4-5 x}+4=-333⋅e4−5x+4=−33 x=x=x= Question Help: Written Example Submit Question

Studdy Solution

STEP 1

STEP 2

1. Isolate the exponential term2. Apply the natural logarithm3. Solve for x

STEP 3

First, we want to get the exponential term all by itself.Let's **subtract** 444 from both sides of the equation: 3⋅e4−5x+4−4=−33−43 \cdot e^{4-5x} + 4 - 4 = -33 - 43⋅e4−5x+4−4=−33−4 3⋅e4−5x=−373 \cdot e^{4-5x} = -373⋅e4−5x=−37

STEP 4

Next, we **divide** both sides by 333: 3⋅e4−5x3=−373 \frac{3 \cdot e^{4-5x}}{3} = \frac{-37}{3} 33⋅e4−5x​=3−37​ e4−5x=−373 e^{4-5x} = -\frac{37}{3} e4−5x=−337​

STEP 5

Now, we'll **apply** the natural logarithm (ln) to both sides.This is the key step that lets us solve for xxx because logarithms are the inverse of exponential functions: ln⁡(e4−5x)=ln⁡(−373) \ln(e^{4-5x}) = \ln\left(-\frac{37}{3}\right) ln(e4−5x)=ln(−337​)

STEP 6

Whoa, hold on!We have a problem here.We're trying to take the logarithm of a **negative number**, −373- \frac{37}{3}−337​.This isn't allowed!The logarithm of a negative number is undefined.

STEP 7

Since we encountered an undefined logarithm, there's **no real solution** for xxx that satisfies the original equation.

STEP 8

x=x = x= DNE

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Studdy solves anything!

Start learning now

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

QuestionUse logarithms to solve. Give an exact simplified answer. Enter DNE if there is no solution. $3 \cdot e^{4-5 x}+4=-33$ $x=$ Question Help: Written Example Submit Question

1. Isolate the exponential term
2. Apply the natural logarithm
3. Solve for x

First, we want to get the exponential term all by itself.
Let's subtract $4$ from both sides of the equation: $3 \cdot e^{4-5x} + 4 - 4 = -33 - 4$ $3 \cdot e^{4-5x} = -37$

Next, we divide both sides by $3$ : $\frac{3 \cdot e^{4-5x}}{3} = \frac{-37}{3}$ $e^{4-5x} = -\frac{37}{3}$

Now, we'll apply the natural logarithm (ln) to both sides.
This is the key step that lets us solve for $x$ because logarithms are the inverse of exponential functions: $\ln(e^{4-5x}) = \ln\left(-\frac{37}{3}\right)$

Whoa, hold on!
We have a problem here.
We're trying to take the logarithm of a negative number, $- \frac{37}{3}$ .
This isn't allowed!
The logarithm of a negative number is undefined.

Since we encountered an undefined logarithm, there's no real solution for $x$ that satisfies the original equation.

$x =$ DNE