# Solve the equations for xx^x=100 and x^x=10,000 answers correct to 3 significant figures. Try to solve it without using logarithm.

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1) x^x=100. to solve for x:

f(x)=x^x is a continous function and 3^3=27 and 4^4=256. therefore, there is a solution beween x= 3 and x=4.

Try around 3+ (4-3)(100-3^3)/(4^4-3^3)=3.3

f(3.3)=51.415, f(3.4)=64.125,f(3.5)=80.211, f(3.6)=100.621.

By conitnuity, f(x)=100 has has solution for x between 3.5 and 3.6.

try for x aound 3.5+(3.6-3.5)(100-80.211)/(100.621-80.211)=3.597

f(3.597)=99.935

f(3.598)=100.163

Thus between f(3.59)=98.353 and f(3.60)=100.621 . f(3.60) is nearest to 100. So, we can conclude the value of x=3.597 or 3.59 for 3 significant places or rounded to 3.60.

ii)

x^x=10000, to solve for x:

f(x)=x^5= 3125 and 6^6=46656. Therefore x^x =10000, has a solution between 5 and 6.

5.4^5.4 =9014.182 and 5.5^5.5=11803.065

Try for x^x for x=5.4+(10000-5.54^5.4)/(5.5^5-5.4^4)

=5.435348.

Therefore, f(x)= 5.43^5.43=9771.541 and f(5.44)=10038.250. Thus we can continue for a better accuracy. But for 3 significant places x=5.44 is the solution for which x^x is nearest to 10000.