Number of Digits Calculation for 6060
To find the number of digits in the expansion of a positive integer \(N\), we use the base-10 logarithm. The number of digits is given by the formula:
\(\text{Number of digits} = \lfloor \log_{10}(N) \rfloor + 1\)
Here, the number \(N\) is \(60^{60}\). So we need to calculate \(\lfloor \log_{10}(60^{60}) \rfloor + 1\).
Logarithm Calculation Steps
First, let's calculate \(\log_{10}(60^{60})\). Using the power rule of logarithms (\(\log(a^b) = b \log(a)\)), we get:
\(\log_{10}(60^{60}) = 60 \times \log_{10}(60)\)
Now, we need to find the value of \(\log_{10}(60)\). We can express \(60\) as a product of numbers whose logarithms are known or easily found:
\(60 = 6 \times 10 = (2 \times 3) \times 10\)
Using the product rule of logarithms (\(\log(abc) = \log(a) + \log(b) + \log(c)\)):
\(\log_{10}(60) = \log_{10}(2 \times 3 \times 10) = \log_{10}(2) + \log_{10}(3) + \log_{10}(10)\)
We are given the values:
- \(\log_{10}(2) = 0.301\)
- \(\log_{10}(3) = 0.477\)
- We know that \(\log_{10}(10) = 1\)
Substituting these values:
\(\log_{10}(60) = 0.301 + 0.477 + 1 = 1.778\)
Now, substitute this back into the expression for \(\log_{10}(60^{60})\):
\(\log_{10}(60^{60}) = 60 \times 1.778\)
Let's perform the multiplication:
\(60 \times 1.778 = 106.68\)
Final Digit Count Determination
Now we apply the formula for the number of digits:
\(\text{Number of digits} = \lfloor \log_{10}(60^{60}) \rfloor + 1\)
\(\text{Number of digits} = \lfloor 106.68 \rfloor + 1\)
The floor of \(106.68\) is \(106\). So:
\(\text{Number of digits} = 106 + 1 = 107\)
Therefore, the number of digits in the expansion of \(60^{60}\) is \(107\).
Similar Questions
If \(a^b = b^a\), then what is
\(\frac{a \times (\frac{a}{b}) ^{ \frac{a}{b} } }{ (a)^{ \frac{a}{b} } }\)
equal to?
- If \(\alpha\) and \(\beta\) are the roots of the equation \(\log_{10} [998+\sqrt{x^2-18x+76}] = 3\) then what is \((\alpha - \beta)^2\) equal to?
If \(x^4 + y^4 = 14x^2 y^2\), then consider the following :
I. \(\log_{10}(x^2 + y^2)\) = \(\log_{10} x + \log_{10} y + 2\log_{10} 2\)
II. \(\log_{10}(x^2-y^2)=\log_{10} x + \log_{10} y\) +\(\log_{10}2+0.5\log_{10} 3\)
Which of the above is/are correct?
If log10(100001 - 4x)/(5 - x) = 1, then what is x equal to?
What is √17 - 4√15 + √8 - 2√15 equal to?
- If \(\log_{10}(80)=p\), \(\log_{10}(45)=q\) and \(\log_{10}(216)=r\), then what is \(\log_{10}(384)\) equal to ?
Consider the following statements in respect of common logarithms :
I. The logarithm of a number greater than 100 but less than 1000 lies between 2 and 3.
II. The logarithm of a positive number less than unity is negative.
Which of the statements given above is/are correct?
What is the number of zeros immediately after the decimal point in \((0.5)^{1000}\)? (Given that \(\log_{10}2 = 0.30103\))
A sum of money at the rate of \(5\%\) per annum compounded annually becomes \(n\) times in \(100\) years. What is the value of \(n\)? (Given \(\log_{10}2=0.301\), \(\log_{10}3=0.477\) and \(\log_{10}7=0.845\))
Important Questions from Logarithms
Given that $132^{0.14} = x$, $132^{0.26} = y$ and $x^z = y^2$, then the value of z is close to:
For \(x \ge y > 1\), let \(\log_x\left(\frac{x}{y}\right) + \log_y\left(\frac{y}{x}\right) = k\), then the value of \(k\) can never be equal to
If \(\log_ba = p\), \(\log_dc = 2p\) and \(\log_fe = 3p\), then what is \((ace)^{\frac{1}{p}}\) equal to ?
- What is the number of solutions of \(\log_4(x-1) = \log_2(x - 3)\)?
- Let \(p = \ln(x)\), \(q = \ln(x^3)\) and \(r = \ln(x^5)\), where \(x > 1\). Which of the following statements is/are correct?
I. \(p, q\) and \(r\) are in AP.
II. \(p, q\) and \(r\) can never be in GP.
Select the answer using the code given below.
