22.44x⁻¹˙⁵+16.8x⁻⁰˙⁰⁵=1.875.
When x=1, the left-hand side becomes 39.24.
Let x=10¹⁰ the left-hand side becomes:
22.44×10⁻¹⁵+1.68√10=5.313 approx. So the left-hand side decreases as x increases. The first term becomes insignificant as x increases. When x=10²⁰, we have:
22.44×10⁻³⁰+1.68=1.68 approx. Therefore the solution is between x=10 and 20.
An approximation for the solution is 16.8x⁻⁰˙⁰⁵=1.875.
x⁻⁰˙⁰⁵=0.1116 approx, -0.05log(x)=log(0.1116),
log(x)=19.0462, x=1.1121×10¹⁹.
This is probably not the answer intended so I suggest the equation should have been:
22.44x⁻¹˙⁵+16.8x⁻⁰˙⁵=1.875,
22.44+16.8x=1.875x¹˙⁵,
(22.44+16.8x)²=1.875²x³,
1.875²x³-16.8²x²-2(22.44)(16.8x)-22.44²=0.
For convenience, let a=1.875, b=16.8, c=22.44:
a²x³-b²x²-2bcx-c²=0.
We can use Newton’s Method to solve this cubic.
First, differentiate: 3a²x²-2b²x-2bc
x=x-(a²x³-b²x²-2bcx-c²)/(3a²x²-2b²x-2bc) is the iterative equation.
Graphically, the only real solution is near x=80, so we use this to start the iterative process:
x₀=80,
x₁=83.11096295,
x₂=82.89096876,
x₃=82.88981613,
x₄=82.8898161,
x₅=82.8898161017.
Reduce this to 4 decimal places: x=82.8898.