Answer :
Let's solve each equation step-by-step to find the value of [tex]\( x \)[/tex]:
### (i) [tex]\(2^x = 4\)[/tex]
First, recall that [tex]\(4\)[/tex] can be written as a power of [tex]\(2\)[/tex]:
[tex]\[4 = 2^2\][/tex]
So, we can rewrite the equation as:
[tex]\[2^x = 2^2\][/tex]
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[x = 2\][/tex]
### (ii) [tex]\(4^x = 64\)[/tex]
Next, recall that [tex]\(64\)[/tex] can be written as a power of [tex]\(4\)[/tex]:
[tex]\[64 = 4^3\][/tex]
So, we can rewrite the equation as:
[tex]\[4^x = 4^3\][/tex]
Again, since the bases are the same, we set the exponents equal to each other:
[tex]\[x = 3\][/tex]
### (iii) [tex]\(8^x = 1\)[/tex]
Recall that any number raised to the power of zero is [tex]\(1\)[/tex]:
[tex]\[8^0 = 1\][/tex]
Therefore, we set the two expressions equal:
[tex]\[x = 0\][/tex]
### (iv) [tex]\(3^x = \frac{1}{3}\)[/tex]
We can rewrite [tex]\(\frac{1}{3}\)[/tex] as [tex]\(3^{-1}\)[/tex]:
[tex]\[3^x = 3^{-1}\][/tex]
Hence, setting the exponents equal, we get:
[tex]\[x = -1\][/tex]
### (v) [tex]\(4^x = \frac{1}{64}\)[/tex]
We can rewrite [tex]\(\frac{1}{64}\)[/tex] as [tex]\(64^{-1}\)[/tex]. Since [tex]\(64 = 4^3\)[/tex], we can write:
[tex]\[\frac{1}{64} = (4^3)^{-1} = 4^{-3}\][/tex]
So the equation becomes:
[tex]\[4^x = 4^{-3}\][/tex]
Setting the exponents equal:
[tex]\[x = -3\][/tex]
### (vi) [tex]\(3^{2x + 4} = 343\)[/tex]
Firstly, observe that [tex]\(343\)[/tex] can be written as a power of [tex]\(7\)[/tex]:
[tex]\[343 = 7^3\][/tex]
This equation is more complicated because the bases are not the same. So we take the natural logarithm ([tex]\(\log\)[/tex]) on both sides to solve for [tex]\(x\)[/tex]:
[tex]\[\ln(3^{2x + 4}) = \ln(7^3)\][/tex]
Using the property of logarithms [tex]\(\ln(a^b) = b\ln(a)\)[/tex], we have:
[tex]\[(2x + 4) \ln(3) = 3 \ln(7)\][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[2x + 4 = \frac{3 \ln(7)}{\ln(3)}\][/tex]
Subtract 4 from both sides:
[tex]\[2x = \frac{3 \ln(7)}{\ln(3)} - 4\][/tex]
Divide by 2:
[tex]\[x = \frac{\frac{3 \ln(7)}{\ln(3)} - 4}{2}\][/tex]
Thus, the value in simplified symbolic form is:
[tex]\[x = \frac{\ln\left(7 \sqrt{7} / 9\right) + \pi i}{\ln(3)}\][/tex]
So finally, the solutions are:
[tex]\[ \begin{align*} (i) & \quad x = 2 \\ (ii) & \quad x = 3 \\ (iii) & \quad x = 0 \\ (iv) & \quad x = -1 \\ (v) & \quad x = -3 \\ (vi) & \quad x = \frac{\ln\left(7 \sqrt{7} / 9\right) + \pi i}{\ln(3)} \end{align*} \][/tex]
These are the values of [tex]\( x \)[/tex] that satisfy each of the given equations.
### (i) [tex]\(2^x = 4\)[/tex]
First, recall that [tex]\(4\)[/tex] can be written as a power of [tex]\(2\)[/tex]:
[tex]\[4 = 2^2\][/tex]
So, we can rewrite the equation as:
[tex]\[2^x = 2^2\][/tex]
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[x = 2\][/tex]
### (ii) [tex]\(4^x = 64\)[/tex]
Next, recall that [tex]\(64\)[/tex] can be written as a power of [tex]\(4\)[/tex]:
[tex]\[64 = 4^3\][/tex]
So, we can rewrite the equation as:
[tex]\[4^x = 4^3\][/tex]
Again, since the bases are the same, we set the exponents equal to each other:
[tex]\[x = 3\][/tex]
### (iii) [tex]\(8^x = 1\)[/tex]
Recall that any number raised to the power of zero is [tex]\(1\)[/tex]:
[tex]\[8^0 = 1\][/tex]
Therefore, we set the two expressions equal:
[tex]\[x = 0\][/tex]
### (iv) [tex]\(3^x = \frac{1}{3}\)[/tex]
We can rewrite [tex]\(\frac{1}{3}\)[/tex] as [tex]\(3^{-1}\)[/tex]:
[tex]\[3^x = 3^{-1}\][/tex]
Hence, setting the exponents equal, we get:
[tex]\[x = -1\][/tex]
### (v) [tex]\(4^x = \frac{1}{64}\)[/tex]
We can rewrite [tex]\(\frac{1}{64}\)[/tex] as [tex]\(64^{-1}\)[/tex]. Since [tex]\(64 = 4^3\)[/tex], we can write:
[tex]\[\frac{1}{64} = (4^3)^{-1} = 4^{-3}\][/tex]
So the equation becomes:
[tex]\[4^x = 4^{-3}\][/tex]
Setting the exponents equal:
[tex]\[x = -3\][/tex]
### (vi) [tex]\(3^{2x + 4} = 343\)[/tex]
Firstly, observe that [tex]\(343\)[/tex] can be written as a power of [tex]\(7\)[/tex]:
[tex]\[343 = 7^3\][/tex]
This equation is more complicated because the bases are not the same. So we take the natural logarithm ([tex]\(\log\)[/tex]) on both sides to solve for [tex]\(x\)[/tex]:
[tex]\[\ln(3^{2x + 4}) = \ln(7^3)\][/tex]
Using the property of logarithms [tex]\(\ln(a^b) = b\ln(a)\)[/tex], we have:
[tex]\[(2x + 4) \ln(3) = 3 \ln(7)\][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[2x + 4 = \frac{3 \ln(7)}{\ln(3)}\][/tex]
Subtract 4 from both sides:
[tex]\[2x = \frac{3 \ln(7)}{\ln(3)} - 4\][/tex]
Divide by 2:
[tex]\[x = \frac{\frac{3 \ln(7)}{\ln(3)} - 4}{2}\][/tex]
Thus, the value in simplified symbolic form is:
[tex]\[x = \frac{\ln\left(7 \sqrt{7} / 9\right) + \pi i}{\ln(3)}\][/tex]
So finally, the solutions are:
[tex]\[ \begin{align*} (i) & \quad x = 2 \\ (ii) & \quad x = 3 \\ (iii) & \quad x = 0 \\ (iv) & \quad x = -1 \\ (v) & \quad x = -3 \\ (vi) & \quad x = \frac{\ln\left(7 \sqrt{7} / 9\right) + \pi i}{\ln(3)} \end{align*} \][/tex]
These are the values of [tex]\( x \)[/tex] that satisfy each of the given equations.
