Solving $5^n > 4,000,000$ without a calculator 
If $n$ is an integer and $5^n > 4,000,000.$ What is the least possible value of $n$? (answer: $10$)

How could I find the value of $n$ without using a calculator ?
 A: I dunno, this is a tough one, especially without a calculator.
Here is the Python program I used to figure this one out:
for n in range(1,11):
    print "5^%s-4,000,000 = %s" % (n, pow(5,n)-4000000)

Here is the output:
5^1-4,000,000 = -3999995.0
5^2-4,000,000 = -3999975.0
5^3-4,000,000 = -3999875.0
5^4-4,000,000 = -3999375.0
5^5-4,000,000 = -3996875.0
5^6-4,000,000 = -3984375.0
5^7-4,000,000 = -3921875.0
5^8-4,000,000 = -3609375.0
5^9-4,000,000 = -2046875.0
5^10-4,000,000 = 5765625.0

It looks like $n=10$ is the answer.
A: It helps if you remember that $ln(2) \approx 0.7$ and $ln(10) \approx 2.3$. (These are common bases to work in, so they're generally useful numbers.)
$$\begin{align}
5^n &> 4\ 000\ 000\\
\ln(5^n) &> \ln(4\ 000\ 000)\\
n (\ln 10 - \ln 2) &> 2 \ln 2 + 6 \ln 10\\
n (2.3 - 0.7) &> 2 \times 0.7 + 6 \times 2.3\\
1.6 n &> 1.4 + 13.8\\
1.6 n &> 15.2\\
1.6 n &> 16 - 0.8\\
n &> 10 - 0.5\\
n &> 9.5\\
n &= 10
\end{align}$$
A bit much for mental arithmetic, but quite doable just typing into this here box.
A: \begin{eqnarray}
&  5^n &>& 4.000.000\\
\Leftrightarrow & 5^n &>& 5^6 \cdot 2^8 \\
\Leftrightarrow & 5^{n-6} &>& 256.\\
\end{eqnarray}
Then, $n=10$.
A: Divide 4000000 by 5, without a calculator, getting 800000. Divide again; 160000. Again; 32000. Then 6400, then 1280, then 256, then 51 (rounding), then 10, then 2. So $2\times5^9$ is about 4000000, so $5^{10}$ exceeds 4000000. 
A: 
The easiest way to multiply by $5$ without a calculator is to multiply
  by $10$ and then divide by $2$, i.e.: $$1: 5\times 5 = 50/2 = 25$$ $$2: 250/2 = 125$$ $$3: 1250/2 = 625$$ $$4: 6250/2 = 3125\ldots$$ Won't take you very long to get to
  $10$.

True. And for large power, use approximations :
$$5^{n} = \frac{10^n}{2^n} $$
$$5^9 = \frac{10^9}{2^9} = \frac{1,000,000,000}{512} \cong \frac{1,000,000,000}{500} \cong 2*10^6 < 4*10^6 $$
$$5^{10} = \frac{10^{10}}{2^{10}} = \frac{10,000,000,000}{1024} \cong \frac{10,000,000,000}{1000} \cong 10^7 > 4*10^6$$
It's not a mathematical way to prove, but it's a way to find the result using approximation.
A: The easiest way to multiply by 5 without a calculator is to multiply by 10 and then divide by 2.  ie:  1: 5x5 = 50/2 = 25.   2: 250/2 = 125.  3: 1250/2 = 625.  4: 6250/2 = 3125...  Won't take you very long to get to 10.
A: $$\log_{10}(5^n)=n\cdot \log_{10}(5)\approx n\cdot 0.7$$
$$\log_{10}(4000000)=\log_{10}(4)+6\approx 6.6$$
$$7\cdot 9=63\ \text{so that }\ \boxed{n=10}\ $$
$\log_{10}(2)\approx 0.3$ was only used giving $\log_{10}(5)=\log_{10}(10)-\log_{10}(2)$ and $\log_{10}(4)=2\cdot\log_{10}(2)$  
(if non integer values are allowed $n\approx \frac {6.60206}{0.69897}$)
A: $4,000,000 = 2^2 \times 10^6 = 2^8 \times 5^6$, so you want $5^{n-6} > 2^8 = 256$.  Well, $5^3 = 125\ldots$.
A: By logarithm rules: 
$$5^{n}>4\cdot10^{6}\iff n>\log_{5}2^{2}2^{6}5^{6}=\log_{5}2^{8}+\log_{5}5^{6}=\log_{5}2^{8}+6=\log_{5}256+6$$
Since these are relatively small numbers I assume it is ok to write
: $5^{3}=125$ thus clearly $3<\log_{5}256<4$ hence the minimal $n$
that satisfies this inequality is $4+6=10$ 
A: $5^n > 4,000,000$ find integer $n$.


*

*take $\log10$ of both sides

*$n\log5>2\log2+6\log10=2\log2+6$

*$n > (2\log2+6)/\log5$  (recall from log paper log2~0.3 and log5~0.7 within a few %.

*$n > 6.6/0.7$ ~ $9.4$ thus rounded up 

*$n = 10$

A: Taking square roots of both sides we solve $5^r>2000=25\cdot80$. The right side is approximated from below by $5^2\cdot5^2\cdot3$ so we want $5^{r-4}>3$ or $r=5$, so $n=2r=10$. Check $n=9$ is too small: $5^9<2^9\cdot 3^9<2^9\cdot 3^5\cdot 3^4=512\cdot 243\cdot 81<125,000\cdot 100<4,000,000$.
