Sure, let's solve the given equation step-by-step:
[tex]\[ 3^{2x - 1} = 9^2 \times 27 \][/tex]
1. Simplify the right-hand side:
- First, notice that [tex]\(9\)[/tex] and [tex]\(27\)[/tex] can be written as powers of 3:
[tex]\[
9 = 3^2 \quad \text{and} \quad 27 = 3^3
\][/tex]
2. Substitute these into the equation:
[tex]\[
9^2 = (3^2)^2 = 3^{2 \cdot 2} = 3^4
\][/tex]
[tex]\[
27 = 3^3
\][/tex]
So, the equation becomes:
[tex]\[
3^{2x - 1} = 3^4 \times 3^3
\][/tex]
3. Combine the exponents on the right-hand side:
[tex]\[
3^{2x - 1} = 3^{4 + 3} = 3^7
\][/tex]
4. Since the bases are the same, we can equate the exponents:
[tex]\[
2x - 1 = 7
\][/tex]
5. Solve for [tex]\( x \)[/tex]:
[tex]\[
2x - 1 = 7
\][/tex]
Add 1 to both sides:
[tex]\[
2x = 8
\][/tex]
Divide by 2:
[tex]\[
x = 4
\][/tex]
So, one of the solutions is [tex]\( x = 4 \)[/tex].
However, we must also consider the possibility of complex solutions arising from logarithmic properties. Solve the equation more generally using logarithms:
6. Rewrite the equation using logarithms for a more general solution:
[tex]\[
3^{2x - 1} = 9^2 \times 27
\][/tex]
With logarithms:
[tex]\[
2x - 1 = \log_3(9^2 \times 27)
\][/tex]
Now calculate:
[tex]\[
\log_3(9^2 \times 27) = \log_3(3^{4+3}) = \log_3 (3^7) = 7
\][/tex]
[tex]\[
2x - 1 = 7
\][/tex]
Hence:
[tex]\[
x = 4
\][/tex]
But considering the logarithmic nature, there is also a complex part which generally appears due to periodicity of complex logarithms.
So, the solutions are:
[tex]\[ x = 4 \][/tex]
And, using properties of logarithms, another solution might appear:
[tex]\[ x = \frac{\log(81) + i\pi}{\log(3)} \][/tex]
Thus, the final solutions are:
[tex]\[ x = 4 \][/tex]
and
[tex]\[ x = \frac{\log(81) + i\pi}{\log(3)} \][/tex]
