How do I solve the equation involving logarithms and exponents for x?

[anemone]

Solve the equation $\left(\dfrac{1}{10}\right)^{\large\log_{0.25}(\sqrt[4]{x}-1)}-4^{\large\log_{10}(\sqrt[4]{x}+5)}=6$ for $x\ge 1$.

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[anemone]

No one answered last week's problem. :( You can find the proposed solution below:

Spoiler

Solve the equation $\left(\dfrac{1}{10}\right)^{\large\log_{0.25}(\sqrt[4]{x}-1)}-4^{\large\log_{10}(\sqrt[4]{x}+5)}=6$ for $x\ge 1$.

Let $a=\log_{0.25}(\sqrt[4]{x}-1)$ and $b=\log_{10}(\sqrt[4]{x}+5)$. Then $(0.25)^a=\sqrt[4]{x}-1$ and $10^b=\sqrt[4]{x}+5$, subtracting these two equations yields $(0.25)^a-10^b=-6$--(*).

On the other hand, the given equation can be expressed as $\left(\dfrac{1}{10}\right)^a-4^b=6$---(**).

Adding (*) and (**) gives

$\left(\dfrac{1}{4}\right)^a-4^b+\left(\dfrac{1}{10}\right)^a-10^b=0$

$(4^{-a}-4^b)+(10^{-a}-10^b)=0$

The left side is less than 0 when $-1<b$ and greater than 0 when $-1>b$. Therefore, $-a=b$ and so $10^b-4^b=6$. One solution of this is $b=1$.

We show that this solution is unique. Observe that the function $f(x)=6\left(\dfrac{1}{10}\right)^x+\left(\dfrac{4}{10}\right)^x$ decreases as $x$ increases from 0 and takes the values 1 when $x=1$. Since $f(x)=1$ is equivalent to $6=10^x-4^x$, we see that $x=1$ is the only solution of the latter equation.