Math  /  Algebra

Question4.) Simplify or solve the following equation, round'to two decimal places if needed'(show your work when possible). a.) ln(e4)\ln \left(e^{4}\right) b.) ln(x)=10\ln (x)=10 c.) e3+xe2+x=40e^{3+x} * e^{2+x}=40

Studdy Solution

STEP 1

What is this asking? We're tackling some logarithmic and exponential equations involving *e* and natural logarithms (ln).
We need to simplify expressions and solve for *x*. Watch out! Remember the special relationship between ln\ln and *e*!
They're inverses, like addition and subtraction.
Also, be careful with exponent rules!

STEP 2

1. Simplify the natural logarithm of an exponential.
2. Solve a logarithmic equation.
3. Solve an exponential equation.

STEP 3

We're asked to simplify ln(e4)\ln(e^4).
Remember, ln\ln is the logarithm with base *e*.
So, ln(e4)\ln(e^4) is asking: *"e* raised to *what power* equals e4e^4?".

STEP 4

Well, that's easy!
The answer is just **4**.
It's like asking "What do I add to 3 to get 3 + 4?".
The answer is 4.
Here, ln\ln and *e* "undo" each other, just like adding and subtracting the same number.
So, ln(e4)=4\ln(e^4) = 4.

STEP 5

We have ln(x)=10\ln(x) = 10.
To solve for *x*, we need to get rid of the ln\ln.

STEP 6

How do we do that?
We use the fact that *e* and ln\ln are inverses.
Applying the exponential function to both sides we get eln(x)=e10e^{\ln(x)} = e^{10}.

STEP 7

Since ee and ln\ln are inverses, they "undo" each other, leaving us with x=e10x = e^{10}.
So, our solution is x=e10x = e^{10}.

STEP 8

We have e3+xe2+x=40e^{3+x} \cdot e^{2+x} = 40.
First, let's simplify the left side using exponent rules.
When multiplying exponential terms with the same base, we *add* the exponents.

STEP 9

Adding the exponents, we get e(3+x)+(2+x)=e5+2xe^{(3+x) + (2+x)} = e^{5 + 2x}.
So, our equation becomes e5+2x=40e^{5+2x} = 40.

STEP 10

Now, we want to get that *x* down from the exponent.
To do that, we'll use the natural logarithm (ln\ln) on both sides: ln(e5+2x)=ln(40)\ln(e^{5+2x}) = \ln(40).

STEP 11

Since ln\ln and *e* are inverses, they "undo" each other, giving us 5+2x=ln(40)5 + 2x = \ln(40).

STEP 12

Now, we just need to isolate *x*.
Subtract 5 from both sides: 2x=ln(40)52x = \ln(40) - 5.

STEP 13

Finally, divide both sides by 2: x=ln(40)52x = \frac{\ln(40) - 5}{2}.

STEP 14

a) ln(e4)=4\ln(e^4) = 4 b) x=e10x = e^{10} c) x=ln(40)52x = \frac{\ln(40) - 5}{2}

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