Solve for $x$
$$5^{\log x}+5x^{\log 5}=3$$
where base of log is $a$, $a>0$ and $a \neq1$
Could someone hint as how to initiate this question? I am not having any idea as how to proceed.
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1Hint: $5^{\log x}=a^{\log 5 \log x}$. – vadim123 Apr 10 '16 at 20:30
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2$5^{\log x}=a^{\log 5\log x}=x^{\log 5}$, so your equation is $x^{\log 5}=\frac{1}{2}$. – user236182 Apr 10 '16 at 20:32
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The thing to remember is $\log b^c = c \log b.$ In your cases, $\log 5^{\log x} = \log x \; \log 5$ and $\log x^{\log 5} = \log 5 \; \log x.$ Put them together, $$ 5^{\log x} = x^{\log 5} $$
Will Jagy
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Rewrite the statement: $$5^{\ln{x}} + 5\cdot5^{\log_{5}{x}\ln{5}} = 5^{\ln{x}} + 5\cdot5^{lnx} = 6\cdot5^{\ln{x}} = 3$$,then $$x = e^{\log_{5}{\frac{1}{2}}}$$
openspace
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