I'm totally stumped with this one. Don't know where to start. Any hint is appreciated.
For every meter a diver descends below the water surface, the light intensity is reduced by 3.5%. At what depth is the light intensity only 25% of that at the surface?
Note that if we are $1$ meter below the water, the intensity of light is $1-0.035=0.965$ the intensity at the surface.
If we submerge another meter, the intensity is $0.965\times 0.965=(0.965)^2$.
Continuing, if we are $x$ meters below the surface, the intensity is $(0.965)^x$.
We need to find $x$ such that $(0.965)^x=0.25$.
Proceeding, we find that at
$$x=\frac{\log(0.25)}{\log(0.965)}$$
the intensity is 25%.
As the diver goes 1 m.. It reduces by $1-0.035$ and similarly.. After 2 m, it is $(1-0.035) (1-0.035)$ So,we have to solve this equation after $x$ m... $$(1-0.035)^x=0.25$$ Solving gives using logarithms $$x\approx38.9111474113$$
