Let phi (x) = x + 2^{log_x 3} - 3^{log_x 2}. | vedclass

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(A) Given the function $\phi (x) = x + 2^{\log_x 3} - 3^{\log_x 2}$.Using the property of logarithms,$a^{\log_b c} = c^{\log_b a}$,we can simplify the

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$\phi (2) = 2$

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(A) Given the function $\phi (x) = x + 2^{\log_x 3} - 3^{\log_x 2}$.Using the property of logarithms,$a^{\log_b c} = c^{\log_b a}$,we can simplify the terms.Consider the term $2^{\log_x 3}$. By the property,$2^{\log_x 3} = 3^{\log_x 2}$.Substituting this into the function: $\phi (x) = x + 3^{\log_x 2} - 3^{\log_x 2} = x$.Now,check the options:For option $A$: $\phi (2) = 2$. This is true.For option $B$: $\phi (1)$ is undefined because the base of the logarithm $x$ must be $x > 0$ and $x 
eq 1$.For option $C$: $\phi (-1.5)$ is undefined because the base of the logarithm must be positive.Therefore,the correct option is $A$.

Let $\phi (x) = x + 2^{\log_x 3} - 3^{\log_x 2}$. Then which of the following is true?

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