Question
OEFENING 4 Los op vir \( x \) : \( 1 \quad 3^{x}-3^{x-2}=24 \) \( 2 \quad 3^{x+1}=3^{x}+\frac{2}{3} \) \( 3 \quad 3^{x}+3^{x+2}=10 \) \( 4 \quad 5^{x+1}-\frac{3}{25}=2.5^{x} \) \( 5 \quad 2^{x}+2^{x}+2^{x}=96 \) \( 6 \quad 12^{x}+12^{x}=288 \) \( 7 \quad 2^{x}+2^{x-2}=5 \) \( 8 \quad 2^{2 x}+4^{x+1}=80 \)
Ask by Lynch Chambers. in South Africa
Feb 03,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
Here are the solutions for each equation:
1. \( x = 3 \)
2. \( x = -1 \)
3. \( x = 0 \)
4. \( x = -2 \)
5. \( x = 5 \)
6. \( x = 2 \)
7. \( x = 2 \)
8. \( x = 2 \)
Solution
Laten we elk van de oefeningen stap voor stap oplossen:
---
**1. \( 3^{x} - 3^{x-2} = 24 \)**
**Oplossing:**
\[
\begin{align*}
3^{x} - 3^{x-2} &= 24 \\
3^{x-2} \cdot 3^{2} - 3^{x-2} &= 24 \quad (\text{Omdat } 3^{x} = 3^{x-2} \cdot 3^{2}) \\
3^{x-2} (9 - 1) &= 24 \\
3^{x-2} \cdot 8 &= 24 \\
3^{x-2} &= 3 \quad (\text{delen door } 8) \\
3^{x-2} &= 3^{1} \\
x - 2 &= 1 \\
x &= 3
\end{align*}
\]
**Antwoord:** \( x = 3 \)
---
**2. \( 3^{x+1} = 3^{x} + \frac{2}{3} \)**
**Oplossing:**
\[
\begin{align*}
3^{x+1} &= 3^{x} + \frac{2}{3} \\
3 \cdot 3^{x} - 3^{x} &= \frac{2}{3} \quad (\text{Omdat } 3^{x+1} = 3 \cdot 3^{x}) \\
2 \cdot 3^{x} &= \frac{2}{3} \\
3^{x} &= \frac{1}{3} \\
3^{x} &= 3^{-1} \\
x &= -1
\end{align*}
\]
**Antwoord:** \( x = -1 \)
---
**3. \( 3^{x} + 3^{x+2} = 10 \)**
**Oplossing:**
\[
\begin{align*}
3^{x} + 3^{x+2} &= 10 \\
3^{x} (1 + 9) &= 10 \quad (\text{Omdat } 3^{x+2} = 9 \cdot 3^{x}) \\
3^{x} \cdot 10 &= 10 \\
3^{x} &= 1 \\
3^{x} &= 3^{0} \\
x &= 0
\end{align*}
\]
**Antwoord:** \( x = 0 \)
---
**4. \( 5^{x+1} - \frac{3}{25} = 2 \cdot 5^{x} \)**
**Oplossing:**
\[
\begin{align*}
5^{x+1} - \frac{3}{25} &= 2 \cdot 5^{x} \\
5 \cdot 5^{x} - 2 \cdot 5^{x} &= \frac{3}{25} \quad (\text{Omdat } 5^{x+1} = 5 \cdot 5^{x}) \\
3 \cdot 5^{x} &= \frac{3}{25} \\
5^{x} &= \frac{1}{25} \\
5^{x} &= 5^{-2} \\
x &= -2
\end{align*}
\]
**Antwoord:** \( x = -2 \)
---
**5. \( 2^{x} + 2^{x} + 2^{x} = 96 \)**
**Oplossing:**
\[
\begin{align*}
3 \cdot 2^{x} &= 96 \\
2^{x} &= 32 \\
2^{x} &= 2^{5} \\
x &= 5
\end{align*}
\]
**Antwoord:** \( x = 5 \)
---
**6. \( 12^{x} + 12^{x} = 288 \)**
**Oplossing:**
\[
\begin{align*}
2 \cdot 12^{x} &= 288 \\
12^{x} &= 144 \\
12^{x} &= 12^{2} \\
x &= 2
\end{align*}
\]
**Antwoord:** \( x = 2 \)
---
**7. \( 2^{x} + 2^{x-2} = 5 \)**
**Oplossing:**
\[
\begin{align*}
2^{x} + 2^{x-2} &= 5 \\
2^{x-2} \cdot 4 + 2^{x-2} &= 5 \quad (\text{Omdat } 2^{x} = 4 \cdot 2^{x-2}) \\
2^{x-2} (4 + 1) &= 5 \\
2^{x-2} \cdot 5 &= 5 \\
2^{x-2} &= 1 \\
2^{x-2} &= 2^{0} \\
x - 2 &= 0 \\
x &= 2
\end{align*}
\]
**Antwoord:** \( x = 2 \)
---
**8. \( 2^{2x} + 4^{x+1} = 80 \)**
**Oplossing:**
\[
\begin{align*}
2^{2x} + 4^{x+1} &= 80 \\
2^{2x} + (2^{2})^{x+1} &= 80 \quad (\text{Omdat } 4 = 2^{2}) \\
2^{2x} + 2^{2x+2} &= 80 \\
2^{2x} (1 + 4) &= 80 \quad (\text{Omdat } 2^{2} = 4) \\
2^{2x} \cdot 5 &= 80 \\
2^{2x} &= 16 \\
2^{2x} &= 2^{4} \\
2x &= 4 \\
x &= 2
\end{align*}
\]
**Antwoord:** \( x = 2 \)
---
**Samenvatting van de oplossingen:**
1. \( x = 3 \)
2. \( x = -1 \)
3. \( x = 0 \)
4. \( x = -2 \)
5. \( x = 5 \)
6. \( x = 2 \)
7. \( x = 2 \)
8. \( x = 2 \)
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Simplify this solution Bonus Knowledge
Om deze exponentiële vergelijkingen op te lossen, gebruiken we vaak logaritmen en de eigenschappen van exponenten. Voor de eerste vergelijking \( 1: 3^{x}-3^{x-2}=24 \), herschik je het naar \( 3^{x} - \frac{3^x}{9} = 24 \). Dit wordt \( \frac{8}{9}3^{x} = 24 \), wat \( 3^{x} = 24 \cdot \frac{9}{8} = 27 \) oplevert, dus \( x=3 \). Voor \( 2: 3^{x+1} = 3^{x} + \frac{2}{3} \), kan je schrijven als \( 3 \cdot 3^{x} = 3^x + \frac{2}{3} \rightarrow (3-1)3^x = \frac{2}{3} \rightarrow 2 \cdot 3^x = \frac{2}{3} \), wat leidt tot \( 3^x = \frac{1}{3} \) en dus \( x = -1 \). Wil je verder met de andere vergelijkingen?
