Sure, let's rewrite each given logarithmic equation as an exponential equation step-by-step.
1. [tex]\(\log_{23}(5) = \frac{1}{2}\)[/tex]
[tex]\[ 23^{\frac{1}{2}} = 5 \][/tex]
2. [tex]\(\log_{3}\left(\frac{1}{81}\right) = -4\)[/tex]
[tex]\[ 3^{-4} = \frac{1}{81} \][/tex]
3. [tex]\(\log_{\frac{4}{3}}\left(\frac{3}{4}\right) = -1\)[/tex]
[tex]\[ \left(\frac{4}{3}\right)^{-1} = \frac{3}{4} \][/tex]
4. [tex]\(\log_{100}(a) = 2\)[/tex]
[tex]\[ 100^2 = a \][/tex]
(Note: It's not clear what [tex]\( a \)[/tex] should be, but assuming [tex]\( \log_{100}(100^2)=2 \)[/tex])
5. [tex]\( \log_{10}(100) = 2 \)[/tex]
[tex]\[ 10^2 = 100 \][/tex]
6. It's not clear what this one is, it might be erroneous:
[tex]\[ (0 \cdot 1) = -1 \quad \text{(does not make sense as it's not a logarithmic equation)} \][/tex]
7. [tex]\( \ln(e) = 1 \)[/tex]
[tex]\[ e^1 = e \][/tex]
8. [tex]\(\ln\left(\frac{1}{\sqrt{e}}\right) = -\frac{1}{2}\)[/tex]
[tex]\[ e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \][/tex]
So here is the rewritten list with the corresponding exponential equations:
1. [tex]\(23^{\frac{1}{2}} = 5\)[/tex]
2. [tex]\(3^{-4} = \frac{1}{81}\)[/tex]
3. [tex]\(\left(\frac{4}{3}\right)^{-1} = \frac{3}{4}\)[/tex]
4. [tex]\(10^2 = 100\)[/tex]
(Correcting the base assumption here if [tex]\( \log_{10}(100) = 2 \)[/tex])
5. [tex]\((0 \cdot 1) = -1\)[/tex] (Unclear expression for logarithmic/exponential form)
6. [tex]\(e^1 = e\)[/tex]
7. [tex]\(e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}}\)[/tex]
