Divide The whole equation by $4^x$ $$1+\left (\frac 32\right)^x=\left(\frac 32\right)^{2x}$$ Solving the quadratic equation gives $$\left (\frac 32\right)^x=\frac{1+\sqrt 5}{2}$$

NOTE: the other value was not considered since exponential is always positive

This gives us the value of $x$ as $$x=\frac{\ln(1+\sqrt 5)-\ln 2}{\ln 3-\ln 2}$$

But the answer given is $\frac{\ln(\sqrt 5-1)-\ln 2}{\ln 2-\ln 3}$

What is going wrong?


1 Answer 1


Note that

$${1+\sqrt 5}=(1+\sqrt 5)\frac{\sqrt 5 -1}{\sqrt 5 -1}=\frac4{\sqrt 5 -1}$$

therefore the two expressions are equivalent, indeed

$$\frac{\ln(1+\sqrt 5)-\ln 2}{\ln 3-\ln 2}=\frac{\ln\left(\frac4{\sqrt 5 -1}\right)-\ln 2}{\ln 3-\ln 2}=\frac{\ln4-\ln(\sqrt 5 -1)-\ln 2}{\ln 3-\ln 2}=$$

$$=\frac{2\ln2-\ln(\sqrt 5 -1)-\ln 2}{\ln 3-\ln 2}=\frac{-\ln(\sqrt 5 -1)+\ln 2}{\ln 3-\ln 2}=\frac{\ln(\sqrt 5-1)-\ln 2}{\ln 2-\ln 3}$$


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