Question

Solve for \[x\] from the following
\[{7^{1 + x}} + {7^{1 - x}} = 50\]
A. \[ \pm 1\]
B. \[ \pm 3\]
C. \[ \pm 2\]
D. None

Answer 1

Hint: Here we will use the below property for adding powers such as that
\[{a^{1 + x}} = a \times {a^x}\]. This means that if two terms are the multiplication form with the same base value, then their powers will be added.

Complete step-by-step solution:
Step 1: For solving the equation
\[{7^{1 + x}} + {7^{1 - x}} = 50\], we can write it as below:
\[ \Rightarrow {7.7^x} + {7.7^{ - x}} = 50\] \[\left( {\because {a^{1 + x}} = a \times {a^x}} \right)\]
We can write the term
\[{7^{ - x}} = \dfrac{1}{{{7^x}}}\] and by replacing it in the above equation, we get:
\[ \Rightarrow {7.7^x} + 7.\dfrac{1}{{{7^x}}} = 50\]
Step 2: By assuming
\[{7^x} = y\], and replacing it in the above equation \[{7.7^x} + 7.\dfrac{1}{{{7^x}}} = 50\], we get:
\[ \Rightarrow 7y + 7.\dfrac{1}{y} = 50\]
By taking
\[y\] as LCM into the LHS side of the above equation we get:
\[ \Rightarrow \dfrac{{7{y^2} + 7}}{y} = 50\]
By bringing \[y\] into the RHS side of the equation and multiplying it with the term
\[50\], we get:
\[ \Rightarrow 7{y^2} + 7 = 50y\]
After bringing
\[50y\] into the LHS side of the above equation for making it a quadratic equation, we get:
\[ \Rightarrow 7{y^2} - 50y + 7 = 0\] ………….. (1)
Step 3: Now we will solve the above quadratic equation (1), by breaking it into factors. We will write the term
\[50y\] into the form of two terms whose subtraction or addition will give us \[50y\] and multiplication will give
\[49{y^2}\], as shown below:
\[ \Rightarrow 7{y^2} - 49y - y + 7 = 0\]
By taking \[7y\] common from the first two terms of the above equation and
\[1\] the last two terms, we get:
\[ \Rightarrow 7y\left( {y - 7} \right) - 1\left( {y - 7} \right) = 0\]
We can write the above equation as below:
\[ \Rightarrow \left( {7y - 1} \right)\left( {y - 7} \right) = 0\]
We will calculate the two possible value \[y\] from the above equation as shown below:
If \[7y - 1 = 0\] then the value of
\[y\] will be as below:
\[ \Rightarrow y = \dfrac{1}{7}\] ………………….. (2)
And if
\[y - 7 = 0\] then the value of \[y\]will be as below:
\[ \Rightarrow y = 7\] ……………………… (3)
Step 4: By substituting the value of \[y = {7^x}\] in the equation (2) and (3), we get:
\[ \Rightarrow {7^x} = \dfrac{1}{7}\] and \[{7^x} = 7\]
We can write
\[7\]as
\[{7^1}\] and by comparing the powers of the same base value, we get the value of
\[x\] as below:
\[ \Rightarrow {7^x} = {7^{ - 1}}\]
By comparing the powers, we get:
\[ \Rightarrow x = - 1\]
And,
\[{7^x} = 7\], we can write it as below:
\[ \Rightarrow {7^x} = {7^1}\]
By comparing the powers, we get:
\[ \Rightarrow x = + 1\]
Therefore, the value of \[x = \pm 1\]

Option A is the correct answer.

Note: Students should remember some basic property of adding or subtracting powers that the base value of two or more terms should be the same. Some examples as shown below for better understanding:
If two terms with the same base value are in the multiplication form then the powers get added.
 \[ \Rightarrow a \times {a^x} = {a^{1 + x}}\]
If two terms with the same base value are in the division form then the powers get subtracted.
 \[ \Rightarrow a \div {a^x} = {a^{1 - x}}\]