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$ {2}^{1-1/x}=\sqrt[x]{3^x+1}/9 $

I tried to solve this equation using Logarithms:

$ \log_2 {2}^{(x-1)/x}=\log_2 \sqrt[x]{3^x+1}/9 $

then

$ (x-1)/x=\log_2 \sqrt[x]{3^x+1}/9 $

from here I don't know what to do as for the second logarithm I don't know how to extract the exponent.

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  • $\begingroup$ Do you mean $$9\cdot 2^{1-\frac{1}{x}}=(3^x+1)^{1/x}$$? $\endgroup$ – Dr. Sonnhard Graubner Nov 4 '18 at 17:58
  • $\begingroup$ A few logarithm properties that might help somehow. Something to play around with in any event. $$log_a(b^n) = n \cdot log_a(b)$$ $$log_a(b/c) = log_a(b) - log_a(c)$$ $\endgroup$ – Eevee Trainer Nov 4 '18 at 17:59
  • $\begingroup$ ok so $ \log_2 \frac{{2}^{(x-1)/x}}{\sqrt[x]{3^x+1}/9} $ $\endgroup$ – Norman Facco Nov 4 '18 at 18:08
  • $\begingroup$ I don't see an obvious solution of the equation. Are you sure it's $3^x+1$? Shouldn't it read $3^{x+1}$? In the latter case the equation is easy to solve. $\endgroup$ – Michael Hoppe Nov 4 '18 at 18:11
  • $\begingroup$ yes Dr. Sonnhard Graubner, I mean that. $\endgroup$ – Norman Facco Nov 4 '18 at 18:11
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There isn’t a good way to solve this. Always recall the following properties of logarithms when attempting to solve equations. (These are some of the important ones.)

$$\log a+\log b = \log (ab)$$

$$\log a-\log b = \log \frac{a}{b}$$

$$\log a^b = b\log a$$

$$\log_{a^b} c = \frac{1}{b} \log c$$

$$\log_a a = 1$$

$$\log_a b = \frac{1}{\log_b a}$$

Using these properties, it becomes apparent that the equation can’t be solved normally.

$${2}^{1-\frac{1}{x}} = \frac{\sqrt[x]{3^x+1}}{9}$$

$$9\cdot {2}^{1-\frac{1}{x}} = \sqrt[x]{3^x+1}$$

$$\ln \big(9\cdot {2}^{1-\frac{1}{x}}\big) = \ln \sqrt[x]{3^x+1}$$

$$\ln \big(9\cdot {2}^{1-\frac{1}{x}}\big) = \ln ({3^x+1})^{\frac{1}{x}}$$

$$\ln 9+\ln {2}^{1-\frac{1}{x}} = \ln ({3^x+1})^{\frac{1}{x}}$$

$$\ln 9+\big(1-\frac{1}{x}\big)\ln {2} = \frac{1}{x} \ln ({3^x+1})$$

$$\ln 9+\ln 2-\frac{1}{x}\ln 2 = \frac{1}{x} \ln ({3^x+1})$$

Had the question had $3^{x+1}$, you would use the property $\log a^b = b\log a$. From there, you would isolate $x$-terms on one side and then solve for it. However, there is a sum. Logarithms don’t work well with those.

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