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Question Number 124985 by Study last updated on 07/Dec/20

Commented by Study last updated on 07/Dec/20

$$helpme$$

Commented by MJS_new last updated on 07/Dec/20

$$\mathrm{ok}{\mathrm{here}}^{\prime }\mathrm{s}\mathrm{some}\mathrm{help}$$$$\mathrm{start}\mathrm{with}$$$$\underset{0}{\stackrel{2}{\int }}{2}^{x}dx$$$$\underset{0}{\stackrel{3}{\int }}{3}^{2^x}{2}^{x}dx$$$$\underset{0}{\stackrel{4}{\int }}{4}^{3^{2^x}}{3}^{2^x}{2}^{x}dx$$$$...$$$$\mathrm{if}\mathrm{you}\mathrm{can}\mathrm{solve}\mathrm{these}\mathrm{you}\mathrm{can}\mathrm{solve}\mathrm{the}$$$$\mathrm{question}.{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{willing}\mathrm{to}\mathrm{answer}\mathrm{weird}$$$$\mathrm{questions}\mathrm{in}\mathrm{detail}\mathrm{for}\mathrm{people}\mathrm{who}\mathrm{are}\mathrm{just}$$$$\mathrm{fascinated}\mathrm{by}\mathrm{huge}\mathrm{numbers}.$$

Commented by $@y@m last updated on 07/Dec/20

I liked your approach. I think all of us should give hint instead of spoon feeding by solving problems step wise.

Answered by Dwaipayan Shikari last updated on 07/Dec/20

$${\int }_0^2{2}^{x}dx=\frac{2^2-2^0}{log\left(2\right)}$$$${\int }_0^3{3}^{2^x}{2}^{x}dx=\frac{1}{log\left(2\right)}{\int }_1^8{3}^{t}dt=\frac{1}{log\left(3\right)log\left(2\right)}(3^8-3)$$$${\int }_0^4{4}^{3^{2^x}}{3}^{2^x}{2}^{x}dx=\frac{1}{log\left(2\right)}{\int }_1^8{4}^{3^t}{3}^{t}dt=\frac{1}{log\left(2\right)log\left(3\right)}{\int }_3^{3^8}{4}^{u}du$$$$=\frac{1}{log\left(2\right)log\left(3\right)log\left(4\right)}(4^{3^8}-4^3)$$$$....$$$$biggerthan\mathbf{Googol}\left({10}^{100}\right)$$

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