Question

what will be the answer of log40 base 1/20- 1)greater than one 2)smaller than one 3)greater than 0 and smaller than 1 4)None of these

Answer 1

Answer:

answer 3 is right....

Step-by-step explanation:

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Answer 2

Answer:

The Correct answer is Option B

Step-by-step explanation:

Before we solve the question given above, we will first see what a logarithm function is. A logarithm is the power in which a number must be raised in order to get some other number. The logarithm is the inverse function to exponentiation. The general form of logarithm is $\({\log _a}b\)$.Now, we are given a logarithm. We will assume that its value is x. Thus,

$\(x = {\log _{\dfrac{1}{{20}}}}40{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. (i)}}\)$

Now, we will use an exponential property here, which says that $\({a^{ - 1}} = \dfrac{1}{a}\)$ Thus, we get:

$\(x = {\log _{{{20}^{ - 1}}}}40\)$

Now, we will use a logarithm identity in the above question:

$\({\log _{{a^2}}}b = \dfrac{1}{2}{\log _a}b\)$

Thus, we get:

$\(x=-log_{20}40x=-log_{20}(20*2)\)$

Now, we will use $\(\log \left( {a \times b} \right) = \log a + \log b\)$in above equation. Thus, we will get:

$\(x=-[log_{20}20+log_{20}2]x=-[1+log_{20}2]$

Now, we will use$\({\log _a}b = \dfrac{1}{{{{\log }_b}a}}\)$ in above question:

$x=-[1+\frac{1}{log_{2}20} ]\\x=-[1+\frac{1}{1log_{2}(4*5)} ]\\x=-[1+\frac{1}{log24+log25} ]\\x=-[1+\frac{1}{ 2+log25}]..................(ii)$

Now, we consider the value of$\({\log _2}5\)$ as y. Thus,

$\(y = {\log _2}5{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. (iii)}}\)$

We use the identity: $\({\log _a}b = \dfrac{{{{\log }_{10}}b}}{{{{\log }_{10}}a}}\).$

Thus, we get:

$y=\frac{log10_{5}}{log10_{2}} \\\\y=\frac{log_{10} \frac{10}{2}}{log_{10} 2}$

Now, we will use the identity:$\(\log \left( {\dfrac{a}{b}} \right) = \log a - \log b\)$

$y=\frac{log10_{10}-log10_{2}}{ log10_{2}}\\\\y=\frac{1-log10_{2}}{log10_{2}}$

The value of $\({\log _{10}}2\)$is 0.3010. Thus, we have:

$y=\frac{1-0.30100}{0.3010}\\\\y=\frac{0.69900}{.3010} \\\\y=2.322.................(iv)$

From (iii) and (iv), we have:

$\({\log _2}5 = 2.322{\rm{ }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. }}{\rm{. (v)}}\)$

From (ii) and (v), we have the following equation:

$x=-[1+\frac{1}{2+2.322} ]\\x=-[1+\frac{1}{4.322} ]\\x=-[1+0.231]\\x=-[1.231]\\x=-1.231...............(vi)$

From (i) and (vi), we can say that:

$log_\frac{1}{20}40=-1.231\\log_\frac{1}{20}40 < 0$

Hence, option B is correct.