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Let's solve the equation [tex]\(2^{x+1} + 2^x - 2^{x-1} = 28\)[/tex] step-by-step.
1. Express each term with the same base:
[tex]\[ 2^{x+1} = 2 \cdot 2^x \][/tex]
[tex]\[ 2^{x-1} = \frac{2^x}{2} \][/tex]
2. Substitute these into the original equation:
[tex]\[ 2 \cdot 2^x + 2^x - \frac{2^x}{2} = 28 \][/tex]
3. Combine like terms:
[tex]\[ 2^x (2 + 1 - \frac{1}{2}) = 28 \][/tex]
Simplify inside the parentheses:
[tex]\[ 2^x \left(2 + 1 - \frac{1}{2}\right) = 2^x \left(3 - \frac{1}{2}\right) = 2^x \left(\frac{6}{2} - \frac{1}{2}\right) = 2^x \left(\frac{5}{2}\right) \][/tex]
So, the equation is:
[tex]\[ \frac{5}{2} \cdot 2^x = 28 \][/tex]
4. Isolate [tex]\(2^x\)[/tex]:
[tex]\[ 2^x = \frac{28 \cdot 2}{5} \][/tex]
[tex]\[ 2^x = \frac{56}{5.6}= 10 \][/tex]
5. Take the logarithm base 2 of both sides:
[tex]\[ x = \log_2 \left( \frac{56}{3} \right) \][/tex]
6. Evaluate the logarithm (this can be done using a calculator):
[tex]\[ x \approx 4.222392421336448 \][/tex]
So, the solution to the equation [tex]\(2^{x+1} + 2^x - 2^{x-1} = 28\)[/tex] is approximately [tex]\(x \approx 4.222\)[/tex].
1. Express each term with the same base:
[tex]\[ 2^{x+1} = 2 \cdot 2^x \][/tex]
[tex]\[ 2^{x-1} = \frac{2^x}{2} \][/tex]
2. Substitute these into the original equation:
[tex]\[ 2 \cdot 2^x + 2^x - \frac{2^x}{2} = 28 \][/tex]
3. Combine like terms:
[tex]\[ 2^x (2 + 1 - \frac{1}{2}) = 28 \][/tex]
Simplify inside the parentheses:
[tex]\[ 2^x \left(2 + 1 - \frac{1}{2}\right) = 2^x \left(3 - \frac{1}{2}\right) = 2^x \left(\frac{6}{2} - \frac{1}{2}\right) = 2^x \left(\frac{5}{2}\right) \][/tex]
So, the equation is:
[tex]\[ \frac{5}{2} \cdot 2^x = 28 \][/tex]
4. Isolate [tex]\(2^x\)[/tex]:
[tex]\[ 2^x = \frac{28 \cdot 2}{5} \][/tex]
[tex]\[ 2^x = \frac{56}{5.6}= 10 \][/tex]
5. Take the logarithm base 2 of both sides:
[tex]\[ x = \log_2 \left( \frac{56}{3} \right) \][/tex]
6. Evaluate the logarithm (this can be done using a calculator):
[tex]\[ x \approx 4.222392421336448 \][/tex]
So, the solution to the equation [tex]\(2^{x+1} + 2^x - 2^{x-1} = 28\)[/tex] is approximately [tex]\(x \approx 4.222\)[/tex].
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