I figured that \begin{align*}7^{x+1} &= 21^{x}\\ 7^{x+1} &= 3^{x} \times 7^{x} \\ 7 &= 3^{x} \end{align*} but I can't go further at the moment.
I want to Find $3^{2x} + 7^{\frac{1}{x}}$ when $7^{x+1} = 21^{x}$
2
$\begingroup$
exponentiation
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1$3^{2x}=49$ and $7^{\frac 1x}=3$ – 2017-01-03
3 Answers
6
Square both sides to reveal that
$$3^{2x}=49$$
$x$ root both sides to reveal that
$$7^{1/x}=3$$
Thus,
$$3^{2x}+7^{1/x}=52$$
2
You have : $7^{\frac{1}{x}} = 3$ and $3^{2x} = 7^2$. Thus $7^{\frac{1}{x}} + 3^{2x} = 52$
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Take the logarithm on both sides.
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0I do not think this actually helps. Did you try working it out this way? – 2017-01-03
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3Logarithms actually do work, but it's a tedious and bruteforce method. – 2017-01-03
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0This can also help, you find that $x = \frac{\log(7)}{\log(3)} $ . – 2017-01-03
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1I don't know if it is so tedious $x = \frac {\log 7}{\log 3} = \log_3 7, \frac1x = \log_7 3 \implies 3^{2\log_3 7} + 7^{\log_7 3} = 7^2 + 3$ – 2017-01-03
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0@DougM Though definitely not recommended. – 2017-01-03
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0This would be a worthy answer if @DougM's comment were to be incorporated into the body of the answer. – 2017-01-04