Certainly! Let's solve the equation step-by-step:
The given equation is:
[tex]\[ 2^x + 2^{x+1} + 2^{x+2} + 2^{x+1} + 2^{x+4} = 62 \][/tex]
1. Combine like terms:
We start by simplifying the equation by grouping the exponential terms. Notice that some terms have the same bases, so we can simplify by combining them:
[tex]\[ 2^x + 2^{x+1} + 2^{x+2} + 2^{x+1} + 2^{x+4} \][/tex]
This can be rewritten as:
[tex]\[ 2^x + 2 \cdot 2^x + 4 \cdot 2^x + 2 \cdot 2^x + 16 \cdot 2^x \][/tex]
2. Factor out the common term [tex]\(2^x\)[/tex]:
[tex]\[ 2^x (1 + 2 + 4 + 2 + 16) \][/tex]
Simplify the constants inside the parentheses:
[tex]\[ 2^x (1 + 2 + 4 + 2 + 16) = 2^x \cdot 25 \][/tex]
3. Simplify the equation:
Now the equation reduces to:
[tex]\[ 25 \cdot 2^x = 62 \][/tex]
4. Isolate [tex]\(2^x\)[/tex]:
Divide both sides of the equation by 25:
[tex]\[ 2^x = \frac{62}{25} \][/tex]
5. Solve for [tex]\(x\)[/tex]:
To solve for [tex]\(x\)[/tex], we take the logarithm base 2 of both sides:
[tex]\[ x = \log_2 \left( \frac{62}{25} \right) \][/tex]
6. Express the solution:
The solution can also be written using the natural logarithm (logarithm with base [tex]\(e\)[/tex]) as:
[tex]\[ x = \frac{\log \left( \frac{62}{25} \right)}{\log 2} \][/tex]
Or more compactly as:
[tex]\[ x = \log_2 \left( \frac{62}{25} \right) \][/tex]
Thus, the final solution to the equation [tex]\( 2^x + 2^{x+1} + 2^{x+2} + 2^{x+1} + 2^{x+4} = 62 \)[/tex] is:
[tex]\[ x = \log_2 \left( \frac{62}{25} \right) \][/tex]
