Answer :
Let's solve each part of the problem step by step using the given function [tex]\( g(x) = e^x \)[/tex].
(a) [tex]\( g(\ln 3) \)[/tex]
We need to evaluate [tex]\( g(\ln 3) \)[/tex].
1. Start with the definition of [tex]\( g(x) \)[/tex]: [tex]\( g(x) = e^x \)[/tex].
2. Substitute [tex]\( x = \ln 3 \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(\ln 3) = e^{\ln 3} \][/tex]
3. Use the property [tex]\( e^{\ln a} = a \)[/tex] for any positive number [tex]\( a \)[/tex]:
[tex]\[ e^{\ln 3} = 3 \][/tex]
Therefore, [tex]\( g(\ln 3) = 3 \)[/tex].
(b) [tex]\( g\left[\ln \left(2^2\right)\right] \)[/tex]
We need to evaluate [tex]\( g\left[\ln \left(2^2\right)\right] \)[/tex].
1. Start with the definition of [tex]\( g(x) \)[/tex]: [tex]\( g(x) = e^x \)[/tex].
2. Substitute [tex]\( x = \ln (2^2) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g\left[\ln \left(2^2\right)\right] = e^{\ln (2^2)} \][/tex]
3. Use the property [tex]\( \ln (a^b) = b \ln a \)[/tex]:
[tex]\[ \ln \left(2^2\right) = 2 \ln 2 \][/tex]
4. Substitute back in:
[tex]\[ g(2 \ln 2) = e^{2 \ln 2} \][/tex]
5. Use the property [tex]\( e^{\ln a} = a \)[/tex] for any positive number [tex]\( a \)[/tex]:
[tex]\[ e^{2 \ln 2} = (e^{\ln 2})^2 = 2^2 = 4 \][/tex]
Therefore, [tex]\( g\left[\ln \left(2^2\right)\right] = 4 \)[/tex].
(c) [tex]\( g\left[\ln \left(\frac{1}{e^6}\right)\right] \)[/tex]
We need to evaluate [tex]\( g\left[\ln \left(\frac{1}{e^6}\right)\right] \)[/tex].
1. Start with the definition of [tex]\( g(x) \)[/tex]: [tex]\( g(x) = e^x \)[/tex].
2. Substitute [tex]\( x = \ln \left(\frac{1}{e^6}\right) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g\left[\ln \left(\frac{1}{e^6}\right)\right] = e^{\ln \left(\frac{1}{e^6}\right)} \][/tex]
3. Use the property [tex]\( \ln \left(\frac{1}{a}\right) = -\ln a \)[/tex]:
[tex]\[ \ln \left(\frac{1}{e^6}\right) = -\ln (e^6) = -6 \][/tex]
4. Substitute back in:
[tex]\[ g(-6) = e^{-6} \][/tex]
Therefore, [tex]\( g\left[\ln \left(\frac{1}{e^6}\right)\right] = e^{-6} \)[/tex].
Substituting these into our answers, we get:
(a) [tex]\( g(\ln 3) = 3 \)[/tex]
(b) [tex]\( g\left[\ln \left(2^2\right)\right] = 4 \)[/tex]
(c) [tex]\( g\left[\ln \left(\frac{1}{e^6}\right)\right] = e^{-6} \)[/tex]
Converting [tex]\( e^{-6} \)[/tex] to its decimal form, we get approximately [tex]\( 0.00247875 \)[/tex].
Therefore:
(a) [tex]\( g(\ln 3) = 3 \)[/tex]
(b) [tex]\( g\left[\ln \left(2^2\right)\right] = 4 \)[/tex]
(c) [tex]\( g\left[\ln \left(\frac{1}{e^6}\right)\right] = 0.00247875 \)[/tex]
(a) [tex]\( g(\ln 3) \)[/tex]
We need to evaluate [tex]\( g(\ln 3) \)[/tex].
1. Start with the definition of [tex]\( g(x) \)[/tex]: [tex]\( g(x) = e^x \)[/tex].
2. Substitute [tex]\( x = \ln 3 \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g(\ln 3) = e^{\ln 3} \][/tex]
3. Use the property [tex]\( e^{\ln a} = a \)[/tex] for any positive number [tex]\( a \)[/tex]:
[tex]\[ e^{\ln 3} = 3 \][/tex]
Therefore, [tex]\( g(\ln 3) = 3 \)[/tex].
(b) [tex]\( g\left[\ln \left(2^2\right)\right] \)[/tex]
We need to evaluate [tex]\( g\left[\ln \left(2^2\right)\right] \)[/tex].
1. Start with the definition of [tex]\( g(x) \)[/tex]: [tex]\( g(x) = e^x \)[/tex].
2. Substitute [tex]\( x = \ln (2^2) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g\left[\ln \left(2^2\right)\right] = e^{\ln (2^2)} \][/tex]
3. Use the property [tex]\( \ln (a^b) = b \ln a \)[/tex]:
[tex]\[ \ln \left(2^2\right) = 2 \ln 2 \][/tex]
4. Substitute back in:
[tex]\[ g(2 \ln 2) = e^{2 \ln 2} \][/tex]
5. Use the property [tex]\( e^{\ln a} = a \)[/tex] for any positive number [tex]\( a \)[/tex]:
[tex]\[ e^{2 \ln 2} = (e^{\ln 2})^2 = 2^2 = 4 \][/tex]
Therefore, [tex]\( g\left[\ln \left(2^2\right)\right] = 4 \)[/tex].
(c) [tex]\( g\left[\ln \left(\frac{1}{e^6}\right)\right] \)[/tex]
We need to evaluate [tex]\( g\left[\ln \left(\frac{1}{e^6}\right)\right] \)[/tex].
1. Start with the definition of [tex]\( g(x) \)[/tex]: [tex]\( g(x) = e^x \)[/tex].
2. Substitute [tex]\( x = \ln \left(\frac{1}{e^6}\right) \)[/tex] into [tex]\( g(x) \)[/tex]:
[tex]\[ g\left[\ln \left(\frac{1}{e^6}\right)\right] = e^{\ln \left(\frac{1}{e^6}\right)} \][/tex]
3. Use the property [tex]\( \ln \left(\frac{1}{a}\right) = -\ln a \)[/tex]:
[tex]\[ \ln \left(\frac{1}{e^6}\right) = -\ln (e^6) = -6 \][/tex]
4. Substitute back in:
[tex]\[ g(-6) = e^{-6} \][/tex]
Therefore, [tex]\( g\left[\ln \left(\frac{1}{e^6}\right)\right] = e^{-6} \)[/tex].
Substituting these into our answers, we get:
(a) [tex]\( g(\ln 3) = 3 \)[/tex]
(b) [tex]\( g\left[\ln \left(2^2\right)\right] = 4 \)[/tex]
(c) [tex]\( g\left[\ln \left(\frac{1}{e^6}\right)\right] = e^{-6} \)[/tex]
Converting [tex]\( e^{-6} \)[/tex] to its decimal form, we get approximately [tex]\( 0.00247875 \)[/tex].
Therefore:
(a) [tex]\( g(\ln 3) = 3 \)[/tex]
(b) [tex]\( g\left[\ln \left(2^2\right)\right] = 4 \)[/tex]
(c) [tex]\( g\left[\ln \left(\frac{1}{e^6}\right)\right] = 0.00247875 \)[/tex]
