Question:
The approximate value of \( \log_{10}99 \) is (Given \( \log_{10}e = 0.4343 \))
The approximate value of \( \log_{10}99 \) is (Given \( \log_{10}e = 0.4343 \))
Show Hint
For numbers close to powers of 10, use logarithmic approximation formulas to quickly estimate values.
Updated On: Jan 26, 2026
- 1.9657
- 1.9857
- 1.9957
- 1.9757
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Express 99 suitably.
\[ \log_{10}99 = \log_{10}(100 - 1) \] Step 2: Use logarithmic approximation.
For small \(x\), \[ \log(1-x) \approx -x \log e \] Thus, \[ \log_{10}99 = \log_{10}100 + \log_{10}\left(1-\frac{1}{100}\right) \] Step 3: Substitute values.
\[ = 2 - \frac{1}{100}\log_{10}e \] \[ = 2 - \frac{1}{100}(0.4343) \] Step 4: Final calculation.
\[ \log_{10}99 \approx 2 - 0.004343 = 1.9957 \]
\[ \log_{10}99 = \log_{10}(100 - 1) \] Step 2: Use logarithmic approximation.
For small \(x\), \[ \log(1-x) \approx -x \log e \] Thus, \[ \log_{10}99 = \log_{10}100 + \log_{10}\left(1-\frac{1}{100}\right) \] Step 3: Substitute values.
\[ = 2 - \frac{1}{100}\log_{10}e \] \[ = 2 - \frac{1}{100}(0.4343) \] Step 4: Final calculation.
\[ \log_{10}99 \approx 2 - 0.004343 = 1.9957 \]
Was this answer helpful?
0
0
Top MHT CET Mathematics Questions
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- If $\int\frac{f\left(x\right)}{log \left(sin\,x\right)}dx = log\left[log\,sin\,x\right]+c$ then $f\left(x\right)=$
- If $\int\limits^{K}_0 \frac{dx}{2 + 18 x^2} = \frac{\pi}{24}$, then the value of K is
- If $\int^{\pi/2}_{0} \log\cos x dx =\frac{\pi}{2} \log\left(\frac{1}{2}\right)$ then $ \int^{\pi/2}_{0} \log\sec x dx = $
- The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0 $ is
View More Questions
Top MHT CET Logarithms Questions
- Find the value of $ \log_2 32 $.
- If the population grows at the rate of 5% per year, the time taken for the population to become double is (Given \( \log 2 = 0.6912 \))
- Radium decomposes at a rate proportional to the amount present. If half the original amount disappears in 1600 years, then the percentage loss in 100 years is
- If \( \sin(y + z - x) \), \( \sin(z + x - y) \), and \( \sin(x + y - z) \) are in A.P., then
- If the population grows at the rate of 5% per year, then the time taken for the population to become double is (Given \(\log 2 = 0.6912\))
View More Questions
Top MHT CET Questions
- During r- DNA technology, which one of the following enzymes is used for cleaving DNA molecule ?
- In non uniform circular motion, the ratio of tangential to radial acceleration is (r = radius of circle, $v =$ speed of the particle, $\alpha =$ angular acceleration)
- The temperature of $32^{\circ}C$ is equivalent to
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- The heat of formation of water is $ 260\, kJ $ . How much $ H_2O $ is decomposed by $ 130\, kJ $ of heat ?
View More Questions