128 = 125 + 3

$5^{n} > 10^{28}$

$5^{n} > (5 \times 2)^{28}$

$5^{n} > 5^{28} \times 2^{28}$

$5^{n-28} > 2^{28}$

Note that $5^{3} < 2^{7}$ .

Have both sides multiplied to the fourth power, then $5^{12} < 2^{28}$ .

Therefore, $5^{13} > 2^{28}$

Since n - 28 = 13, then n = 41

We simply apply the logarithmic method.

so,

5^n > 10^28

(log both the sides)

= log (5^n) > log (10^28)

= n (log 5) > 28

= n > 28 / (log 5)

= n > 40.0589...

Therefore, n = 41 is the minimum

Taking log 10 of both sides, we have $n \log 5>28 \log 10=28 \Rightarrow n>\frac{28}{\log 5}$ . Since $\log 5$ is just nearly under $\frac{7}{10}$ , we have $n>40.some$ , so the least integer that satisfies this is $\lceil 40.some \rceil$ , which is creary (hopefully?) $\boxed{41}$

(Note that an exacter (?) value of $\log 5$ can be found using a calculator, and that it is a good idea to memorise that $\log 5$ is slightly under $\frac{7}{10}$ (if you haven't already))

For 5^n > 10^ 28, we require n log 5 > 28( log 10 ) > 28 * 1 therefore n > 28/log 5 > 28/.7 > 40 Therefore smallest n = 41

5^n>〖10〗^28 5^n>2^28.5^28 5^n/5^28 >2^28 5^(n-28)>〖〖(2〗^2)〗^14 5^(n-28)>4^14 Because 5^13>4^14. So n – 28 = 13, Thus n = 13 + 28 = 41. Answer : 41

By taking log on both the sides, we get

n log5 > 28 log10

n*log5 > 28

n > 28 / log5

n > 40.2

Hence n = 41

Let 5^n=10^28 then take Lin for both sides .. you will get n * lin 5=28 * lin 10 then n = 40.05894363 add (0.94105637) to get the integer of 41

just added all digits found above of the question "128=125+3". Its 1+2+8+1+2+5+3=22. Then, just count from the top to bottom until reaching 22. Voila, click the answer and got it. :)

5^{n} > 10^{28) Take log for both sides

nlog5 > 28log10 divide by log5 (log10 = 1)

n > 28/{log5} which is approximately equals 40

then the least integer value of n = 41

Take the $\log$ of both sides to the base $10$ . We have $n \log (5) > 28 \Rightarrow n > \dfrac{28}{\log(5)} \Rightarrow n > 40.05\cdots$ . The least integer satisfying this is $\boxed{41}$ .

Taking 28th root and log to the base 10 on both sides we get n>28/log5. log5 is approx. 0.69 hence n>40..

Well, obviously $5^{n} > 10^{28} \Rightarrow 5^{n-28} > 2^{28}$ . Now, if we apply the natural log function on both sides we get

$n = \left \lceil 28 + \frac{\ln2^{28}}{\ln5} \right \rceil = 41$

Take log to base 10 on both sides, thus n*0.6990>28. With a reasonable degree of approximation we can see n=41 or 40 which can be checked in the event of uncertainty yielding n=41