1. log₃0.1+2log₃x=log₃2+log₃5
2. logᵪ5+log₃x=log₃10
3. log[b]9=2
4. ¼log₂16+½logᵪ49=log₂x
1. log₃(0.1x²)=log₃10, so 0.1x²=10, x²=100, and x=10.
2. 1/log₅x=log₃(10/x)=log₅(10/x)/log₅3.
Let y=log₅x, 1/y=(log₅10-log₅x)/log₅3=(log₅10-y)/log₅3.
Cross-multiply: log₅3=ylog₅10-y², y²-ylog₅10+log₅3=0.
If we use the formula to solve the quadratic in y, we discover that the discriminant is negative, because 4log₅3>(log₅10)², so the solution is complex. There is no real solution for x.
3. Rewrite b²=9 so b=3.
4. We can evaluate and simplify: 1+logᵪ7=log₂x. logᵪ7=log₂(x/2).
Therefore 1/log₇x=log₂(x/2), and since log₇x=log₂x/log₂7, log₂7/log₂x=log₂x-1.
Let y=log₂x, log₂7/y=y-1, so y²-y-log₂7=0.
From this, using the quadratic formula y=(1±√(1+4log₂7))/2=2.2485 or -1.2485 approx.
x=2^y=4.7520 or 0.4209 approx.