Find the value of $$5^{\sqrt{\log_57}}-7^{\sqrt{\log_75}}$$ In what form am I supposed to use the identity $a^{\log_nm}=m^{\log_na}$?
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But that would be different, isn't it? $\log_57^{\frac12}=\frac12\log_57\ne\sqrt{\log_57}$ – Tejas Dec 19 '13 at 10:09
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No need to use that identity. You can observe that calling $x = 5^{\sqrt{log_57}}$, then you have $x^{\sqrt{log_57}} = 7$ thus $x = 7^{1/\sqrt{log_57}}$. Then use just the base change formula ($log_ab = log_cb/log_ca $ for $a,b,c > 0$ and $c\neq 1$) to show that $log_57 = 1/log_75$. Your result should then be $0$.
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$a=5^{\sqrt{\log_57}},b=7^{\sqrt{\log_75}}$
$$\log_7a=\log_7(5^{\sqrt{\log_57}})=\sqrt{\log_57}\cdot\log_75=\sqrt{\log_57}\cdot\frac{1}{\log_57}=\frac{1}{\sqrt{\log_57}}$$ $$\log_7b=\log_77^{\sqrt{\log_75}}=\sqrt{\log_75}=\frac{1}{\sqrt{\log_57}}$$
Finally $a=b$
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