Math  /  Algebra

Question8. 2log(8x+4)+6=102 \log (8 x+4)+6=10
9. Assume you deposit $10,000\$ 10,000 into an account paying 4%4 \% annual interest compounds quarterly. How much money will be in your account at the end of 10 years?
10. You want to save $100,000\$ 100,000 for your son's college cost. Assume your account pays 10%10 \% annual interest rate compounded continuously and you invest $50,000now\$ 50,000 \mathrm{now}. How many years will it take to reach your goal?

Studdy Solution

STEP 1

What is this asking? We're solving three different problems: a logarithmic equation, a compound interest problem where we calculate the future value, and another compound interest problem where we find how long it takes to reach a savings goal. Watch out! Remember the logarithm rules and the different compound interest formulas!
Don't mix them up!

STEP 2

1. Solve the Logarithmic Equation
2. Calculate Future Value with Quarterly Compounding
3. Determine Time to Reach Savings Goal with Continuous Compounding

STEP 3

Alright, let's **isolate** that logarithm!
We start with 2log(8x+4)+6=102 \log (8x + 4) + 6 = 10.
Subtract **6** from both sides: 2log(8x+4)=106=42 \log (8x + 4) = 10 - 6 = 4.

STEP 4

Now, **divide** both sides by **2**: log(8x+4)=42=2\log (8x + 4) = \frac{4}{2} = 2.
Remember, we're dividing by 2 to make the coefficient of the logarithm equal to 1.

STEP 5

Time to **rewrite** this logarithm in exponential form!
Since we have a common log (base 10), we get 102=8x+410^2 = 8x + 4, or 100=8x+4100 = 8x + 4.

STEP 6

Let's **isolate** xx.
Subtract **4** from both sides: 1004=8x100 - 4 = 8x, so 96=8x96 = 8x.

STEP 7

Finally, **divide** both sides by **8** to get xx: x=968=12x = \frac{96}{8} = 12.
Boom! x=12x = \textbf{12}!

STEP 8

We're depositing $10,000\$10,000 at 4%4\% annual interest compounded quarterly for 1010 years.
The formula for compound interest is A=P(1+rn)ntA = P(1 + \frac{r}{n})^{nt}, where AA is the **future value**, PP is the **principal** ($10,000\$10,000), rr is the **annual interest rate** (0.040.04), nn is the **number of times interest is compounded per year** (44), and tt is the **time in years** (1010).

STEP 9

Let's **plug** in the values: A=10000(1+0.044)410A = 10000(1 + \frac{0.04}{4})^{4 \cdot 10}.

STEP 10

**Simplify** inside the parentheses: A=10000(1+0.01)40A = 10000(1 + 0.01)^{40}.

STEP 11

**Calculate**: A=10000(1.01)40100001.48886A = 10000(1.01)^{40} \approx 10000 \cdot 1.48886.

STEP 12

So, A14888.64A \approx \textbf{14888.64}.
After 1010 years, you'll have approximately $14,888.64\$\textbf{14,888.64}!

STEP 13

We want to save $100,000\$100,000, starting with $50,000\$50,000, at a 10%10\% annual interest rate compounded continuously.
The formula is A=PertA = Pe^{rt}, where AA is the **future value** ($100,000\$100,000), PP is the **principal** ($50,000\$50,000), rr is the **interest rate** (0.100.10), and tt is the **time** we're solving for.

STEP 14

**Substitute** the values: 100000=50000e0.10t100000 = 50000e^{0.10t}.

STEP 15

**Divide** both sides by 5000050000: 10000050000=e0.10t\frac{100000}{50000} = e^{0.10t}, which simplifies to 2=e0.10t2 = e^{0.10t}.

STEP 16

Take the **natural logarithm** of both sides: ln(2)=ln(e0.10t)\ln(2) = \ln(e^{0.10t}).
This simplifies to ln(2)=0.10t\ln(2) = 0.10t.

STEP 17

**Isolate** tt by dividing by 0.100.10: t=ln(2)0.100.69310.10t = \frac{\ln(2)}{0.10} \approx \frac{0.6931}{0.10}.

STEP 18

So, t6.931t \approx \textbf{6.931} years.
It will take approximately 6.931\textbf{6.931} years to reach your savings goal.

STEP 19

Problem 8: x=12x = 12. Problem 9: $14,888.64\$14,888.64. Problem 10: 6.9316.931 years.

Was this helpful?

Studdy solves anything!

banner

Start learning now

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

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord