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Solve this equation: $$3^{\log_{4}x+\frac{1}{2}}+3^{\log_{4}x-\frac{1}{2}}=\sqrt{x}\qquad (1)$$

I tried to make both sides of the equation have a same base and I started:

$$(1)\Leftrightarrow 3^{\log_{4}x}.\sqrt{3}+ \frac{3^{\log_{4}x}}{\sqrt{3}} = \sqrt{x}$$ $$\Leftrightarrow 3^{\log_{4}x}.3+ 3^{\log_{4}x} = \sqrt{3x}$$ $$\Leftrightarrow 4.3^{\log_{4}x}= \sqrt{3x}$$

At this step, I can't continue. Please help me!

Robert Z
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Trần Tuấn
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5 Answers5

5

Guide:

Taking $\log_4$ on both sides,

$$1 + \log_4 3 \cdot \log_4 x = \frac12 (\log_43 + \log_4 x)$$

Solve for $\log_4 x$.

Siong Thye Goh
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First of all impose the necessary existence conditions, that is: $x > 0$ for the logarithms and $x \geq 0$ for the square root. That is, eventually,

$$x > 0$$

for the whole equation.

Then follow Siong Thye Goh reasoning, obtaining the final equation he wrote.

At that point:

$$1 + \log_4(3)\log_4(x) = \frac{\log_4(3)}{2} + \frac{\log_4(x)}{2}$$

$$\log_4(x)\left(\log_4(3) - \frac{1}{2}\right) = \frac{\log_4(3)}{2} - 1$$

$$\log_4(x) = \frac{\frac{\log_4(3)}{2} - 1}{\log_4(3) - \frac{1}{2}} = \frac{\log_4(3)-2}{2\log_4(3)-1}$$

To solve for $x$ take the exponential base 4 of both terms, getting:

$$\large x = \large 4^{\frac{\log_4(3)-2}{2\log_4(3)-1}}$$

Enrico M.
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Same solution with same approach But simplified with substitution Solution

user579689
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You may also continue as follows:

$$\begin{eqnarray*} 4 \cdot 3^{2 \cdot \log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\ 4 \cdot 9^{\log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\ 4 \cdot 4^{\log_4{9} \cdot \log_{4}\sqrt{x}} &= & \sqrt{3}\sqrt{x} \Leftrightarrow \\ (\sqrt{x})^{\log_4{9} -1} &= & \frac{\sqrt{3}}{4} \Leftrightarrow\\ x & = & \left( \frac{\sqrt{3}}{4}\right)^{\frac{2}{\log_4{9}-1}} \approx 0.0571725 \end{eqnarray*}$$

trancelocation
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Let's generalise a bit, with parameters say $a, b \in (0, \infty)\setminus \{1\}$ subject to $a^2 \neq b$ and let's try to solve the equation:

$$a^{\log_{b}x+\frac{1}{2}}+a^{\log_{b}x-\frac{1}{2}}=\sqrt{x}$$

Notice that the left-hand side can be rewritten as

$$a^{\log_{b}x}(\sqrt{a}+\frac{1}{\sqrt{a}})=b^{\log_{b}a\cdot \log_{b}x}(\sqrt{a}+\frac{1}{\sqrt{a}})=x^{\log_{b}a}(\sqrt{a}+\frac{1}{\sqrt{a}})$$

As you are dealing exclusively with strictly positive reals, your given equation is equivalent to its square, so to speak:

$$\frac{(a+1)^2}{a}x^{2\log_{b}a}=x$$

which leads to

$$x^{2\log_{b}a-1}=\frac{a}{(a+1)^2}$$

Since the right-hand side is never $1$ (you can try to see why), this is why we initially imposed the relation of inequality between $a$ and $b$; it is satisfied in the particular case of your equation.

We finally have the solution:

$$x=\left(\frac{a}{(a+1)^2}\right)^{\frac{1}{2\log_{b}a-1}}$$

ΑΘΩ
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