Sure, let's solve each equation step-by-step to find the value of [tex]\( x \)[/tex]:
### (i) [tex]\( 2^x = 4 \)[/tex]
First, express 4 as a power of 2:
[tex]\[ 4 = 2^2 \][/tex]
So we have:
[tex]\[ 2^x = 2^2 \][/tex]
Since the bases are the same, we can equate the exponents:
[tex]\[ x = 2 \][/tex]
Thus, the value of [tex]\( x \)[/tex] is:
[tex]\[ x = 2 \][/tex]
### (ii) [tex]\( 4^x = 64 \)[/tex]
First, express 64 as a power of 4:
[tex]\[ 64 = 4^3 \][/tex]
So we have:
[tex]\[ 4^x = 4^3 \][/tex]
Again, since the bases are the same, we can equate the exponents:
[tex]\[ x = 3 \][/tex]
Thus, the value of [tex]\( x \)[/tex] is:
[tex]\[ x = 3 \][/tex]
### (iii) [tex]\( 8^x = 1 \)[/tex]
Recall that any non-zero number raised to the power 0 is 1:
[tex]\[ 1 = 8^0 \][/tex]
So we have:
[tex]\[ 8^x = 8^0 \][/tex]
Thus, equating the exponents gives:
[tex]\[ x = 0 \][/tex]
Thus, the value of [tex]\( x \)[/tex] is:
[tex]\[ x = 0 \][/tex]
### (iv) [tex]\( 3^x = \frac{1}{3} \)[/tex]
Express [tex]\(\frac{1}{3}\)[/tex] as a power of 3:
[tex]\[ \frac{1}{3} = 3^{-1} \][/tex]
So we have:
[tex]\[ 3^x = 3^{-1} \][/tex]
Equating the exponents:
[tex]\[ x = -1 \][/tex]
Thus, the value of [tex]\( x \)[/tex] is:
[tex]\[ x = -1 \][/tex]
### (v) [tex]\( 4^x = \frac{1}{64} \)[/tex]
First, express [tex]\(\frac{1}{64}\)[/tex] as a power of 4:
[tex]\[ \frac{1}{64} = 4^{-3} \][/tex]
So we have:
[tex]\[ 4^x = 4^{-3} \][/tex]
Thus, equating the exponents:
[tex]\[ x = -3 \][/tex]
Thus, the value of [tex]\( x \)[/tex] is:
[tex]\[ x = -3 \][/tex]
### (vi) [tex]\( 3^{2x + 4} = 343 \)[/tex]
Express 343 as a power of 7:
[tex]\[ 343 = 7^3 \][/tex]
Rewrite the equation:
[tex]\[ 3^{2x + 4} = 7^3 \][/tex]
To solve for [tex]\( x \)[/tex], we can take the logarithm of both sides with base 3:
[tex]\[ 2x + 4 = \log_3(7^3) \][/tex]
[tex]\[ 2x + 4 = 3 \log_3(7) \][/tex]
Isolate [tex]\( x \)[/tex]:
[tex]\[ 2x = 3 \log_3(7) - 4 \][/tex]
[tex]\[ x = \frac{3 \log_3(7) - 4}{2} \][/tex]
Now, using the change of base formula for logarithms, we have:
[tex]\[ \log_3(7) = \frac{\ln(7)}{\ln(3)} \][/tex]
Substitute this into the equation:
[tex]\[ x = \frac{3 \cdot \frac{\ln(7)}{\ln(3)} - 4}{2} \][/tex]
Thus, the value of [tex]\( x \)[/tex] is approximately:
[tex]\[ x \approx 0.6568656237421333 \][/tex]
So, the final values of [tex]\( x \)[/tex] are:
1. [tex]\( x = 2 \)[/tex]
2. [tex]\( x = 3 \)[/tex]
3. [tex]\( x = 0 \)[/tex]
4. [tex]\( x = -1 \)[/tex]
5. [tex]\( x = -3 \)[/tex]
6. [tex]\( x \approx 0.6568656237421333 \)[/tex]
