Sure, let's solve each part step-by-step.
Given [tex]\( g(x) = e^x \)[/tex], we need to evaluate the following:
### Part (a)
(a) [tex]\( g(\ln 6) \)[/tex]
[tex]\[ g(\ln 6) = e^{\ln 6} \][/tex]
Using the property of exponents and logarithms, [tex]\( e^{\ln a} = a \)[/tex], we get:
[tex]\[ e^{\ln 6} = 6 \][/tex]
So,
[tex]\[ g(\ln 6) = 6 \][/tex]
### Part (b)
(b) [tex]\( g\left[\ln \left(7^2\right)\right] \)[/tex]
[tex]\[ g\left[\ln \left(7^2\right)\right] = e^{\ln (7^2)} \][/tex]
Using the logarithm power rule, [tex]\( \ln (a^b) = b \ln a \)[/tex], we get:
[tex]\[ \ln (7^2) = 2 \ln 7 \][/tex]
Thus,
[tex]\[ e^{\ln (7^2)} = e^{2 \ln 7} \][/tex]
Using the property of exponents and logarithms again:
[tex]\[ e^{2 \ln 7} = (e^{\ln 7})^2 \][/tex]
[tex]\[ (e^{\ln 7})^2 = 7^2 = 49 \][/tex]
So,
[tex]\[ g\left[\ln \left(7^2\right)\right] = 49 \][/tex]
### Part (c)
(c) [tex]\( g\left[\ln \left(\frac{1}{e^5}\right)\right] \)[/tex]
[tex]\[ g\left[\ln \left(\frac{1}{e^5}\right)\right] = e^{\ln (1/e^5)} \][/tex]
Using the property [tex]\( \ln (1/a) = -\ln a \)[/tex]:
[tex]\[ \ln \left(\frac{1}{e^5}\right) = \ln (e^{-5}) = -5 \][/tex]
Thus,
[tex]\[ e^{\ln (1/e^5)} = e^{-5} \][/tex]
So,
[tex]\[ g\left[\ln \left(\frac{1}{e^5}\right)\right] = \frac{1}{e^5} \][/tex]
In summary:
(a) [tex]\( g(\ln 6) = 6 \)[/tex]
(b) [tex]\( g\left[\ln \left(7^2\right)\right] = 49 \)[/tex]
(c) [tex]\( g\left[\ln \left(\frac{1}{e^5}\right)\right] = \frac{1}{e^5} \)[/tex]
