Solve log10(log10(x + 0.6)) = 8.8289 - 3.2511 log10(284 + 273.15) for x.

Step-by-step solution

1. 1

   Collapse the right-hand side to a single number first. It contains no $x$, so it is just arithmetic. Start inside the logarithm:

   $284+273.15=557.15$

   (the shape of adding $273.15$ to a Celsius temperature is the giveaway that this is an Antoine-style vapour-pressure formula).

2. 2

   Evaluate the numeric right side.

   $\log_{10}(557.15)=2.7459721,\qquad3.2511\times2.7459721=8.9274300$

   so

   $8.8289-8.9274300=-0.0985300$

   The result is negative, which is fine: it is the logarithm of a number between $0$ and $1$, not an impossibility.

3. 3

   Undo the outer logarithm. Raise $10$ to the power of each side:

   $\log_{10}(x+0.6)=10^{-0.0985300}=0.7969777$

   Exponentiating turns the negative value into a positive one less than $1$ ✓.

4. 4

   Undo the inner logarithm. Exponentiate once more:

   $x+0.6=10^{0.7969777}=6.2664477$

5. 5

   Solve for $x$ and check the domain.

   $x=6.2664477-0.6=5.6664477\approx5.666$

   Both logarithms are legal at this value: $x+0.6=6.266>0$, and $\log_{10}(6.266)=0.797>0$, so the outer logarithm has a positive argument ✓ — a nested logarithm needs both conditions, and here they hold.

6. 6

   Verify by substituting back. Computing $\log_{10}\left(\log_{10}(5.6664477+0.6)\right)$ returns $-0.0985300$, matching the right-hand side to within $10^{-12}$ ✓. Because the outer $10^{u}$ amplifies errors, carrying at least seven digits through the intermediate steps matters: rounding $-0.0985$ too early shifts $x$ in the third decimal.