Question

practice changing between exponential form and logarithmic form using the table below:
note: \log_b (a) can be written using an underscore like this: \log_b(a).
\

$$\begin{array}{|c|c|} \\hline b^c = a & \\log_b (a) = c \\\\ \\hline & \\log_5 (25) = 2 \\\\ \\hline 8^4 = 4096 & \\\\ \\hline 3^x = 81 & \\\\ \\hline & \\log_4 (y) = 6 \\\\ \\hline 367 = 4^x & \\\\ \\hline w = k^{125} & \\\\ \\hline burnfuzzle = gardyloo^{hipbanana} & \\\\ \\hline \\end{array}$$

question help: \boxed{video}

Explanation:

Step1: Recall the logarithmic-exponential relationship

The definition of a logarithm states that if $\log_{b}(a) = c$, then $b^{c}=a$. We will use this relationship to convert between logarithmic and exponential forms for each row.

Step2: Solve $\log_{5}(25) = 2$ to exponential form

Using the formula $\log_{b}(a)=c\implies b^{c}=a$, here $b = 5$, $a=25$, $c = 2$. So the exponential form is $5^{2}=25$.

Step3: Solve $8^{4}=4096$ to logarithmic form

Using $b^{c}=a\implies\log_{b}(a)=c$, here $b = 8$, $a = 4096$, $c=4$. So the logarithmic form is $\log_{8}(4096)=4$.

Step4: Solve $3^{x}=81$ to logarithmic form

Using $b^{c}=a\implies\log_{b}(a)=c$, here $b = 3$, $a = 81$, $c=x$. So the logarithmic form is $\log_{3}(81)=x$.

Step5: Solve $\log_{4}(y)=6$ to exponential form

Using $\log_{b}(a)=c\implies b^{c}=a$, here $b = 4$, $a = y$, $c = 6$. So the exponential form is $4^{6}=y$, and $4^{6}=4096$, so $y = 4096$.

Step6: Solve $367 = 4^{x}$ to logarithmic form

Using $b^{c}=a\implies\log_{b}(a)=c$, here $b = 4$, $a = 367$, $c=x$. So the logarithmic form is $\log_{4}(367)=x$.

Step7: Solve $w=k^{125}$ to logarithmic form

Using $b^{c}=a\implies\log_{b}(a)=c$, here $b = k$, $a = w$, $c = 125$. So the logarithmic form is $\log_{k}(w)=125$.

Step8: Solve $\text{burnfuzzle}=\text{gardyl oo}_{\text{hipbanana}}$ (assuming it's a symbolic conversion)

Using $b^{c}=a\implies\log_{b}(a)=c$, here $b=\text{gardyl oo}_{\text{hipbanana}}$, $a=\text{burnfuzzle}$, $c$ (exponent) would be such that $(\text{gardyl oo}_{\text{hipbanana}})^{c}=\text{burnfuzzle}$, so the logarithmic form is $\log_{\text{gardyl oo}_{\text{hipbanana}}}(\text{burnfuzzle})=c$ (where $c$ is the exponent from the exponential form, but since it's symbolic, we just apply the formula). And the exponential form from the logarithmic would be $(\text{gardyl oo}_{\text{hipbanana}})^{c}=\text{burnfuzzle}$.

Answer:

• $\log_{5}(25) = 2\implies\boldsymbol{5^{2}=25}$
• $8^{4}=4096\implies\boldsymbol{\log_{8}(4096)=4}$
• $3^{x}=81\implies\boldsymbol{\log_{3}(81)=x}$
• $\log_{4}(y)=6\implies\boldsymbol{4^{6}=y\ (y = 4096)}$
• $367 = 4^{x}\implies\boldsymbol{\log_{4}(367)=x}$
• $w=k^{125}\implies\boldsymbol{\log_{k}(w)=125}$
• $\text{burnfuzzle}=\text{gardyl oo}_{\text{hipbanana}}^{c}\implies\boldsymbol{\log_{\text{gardyl oo}_{\text{hipbanana}}}(\text{burnfuzzle})=c}$ (and vice - versa for exponential form)