Answer
(a) $25^{\log_53x^2}=9x^4$
(b) $\log_e(e^x)=x$
(c) $\log_4(2^{e^x\sin x})=\frac{1}{2}e^x\sin x$
Work Step by Step
(a) $$25^{\log_53x^2}$$
Here $25$ and $5$ do not match each other to use the inverse properties.
But $25=5^2$
That means $$25^{\log_53x^2}=5^{2\log_53x^2}$$
- Apply Power Rule, we have: $2\log_53x^2=\log_5(3x^2)^2=\log_59x^4$
Thus, $$25^{\log_53x^2}=5^{\log_59x^4}=9x^4$$
(b) $$\log_e(e^x)$$
We can apply right away the inverse properties here:
$$\log_e(e^x)=x$$
(c) $$\log_4(2^{e^x\sin x})$$
Notice that $2=\sqrt4=4^{\frac{1}{2}}$
So we can rewrite $2^{e^x\sin x}=4^{\frac{1}{2}e^x\sin x}$
Therefore, $$\log_4(2^{e^x\sin x})=\log_44^{\frac{1}{2}e^x\sin x}$$
Now apply the inverse property with base $a=4$:
$$\log_4(2^{e^x\sin x})=\frac{1}{2}e^x\sin x$$
