Logarithm Inequality

Note that $x>1$ , therefore $t\mapsto \log_x t$ is strictly increasing on $(0,\infty)$ .

then we have : $4\log_x2+\log_x 3=\log_x 48<2 \log_x 7 <\log_x50=2\log_x5+\log_x 2.$

Use the inequalities to get that : $0.45 =\frac{4\times 0.161+0.256}{2} <\log_x 7 < \frac{2\times 0.376 +0.162}{2}= 0.457.$

From here we see that : $\lfloor \log_x 7^{100} \rfloor = \lfloor 100\log_x 7 \rfloor =\boxed{45}.$

I found a straight solution.

If we have $a<\log_x b < c$ , with $a,c>$ and $b>1$ , then it implies that $b^{1/c} < x < b^{1/a}$ .

Then, we have $0.161 < \log_x 2 <0.162 \implies 72.147 < x <74.089$

$0.256 < \log_x 3 <0.257 \implies 71.863 < x <73.073$

$0.375 < \log_x 5 <0.376 \implies 72.271< x<73.100$

Combining the three inequalities, we get,

$72.271 < x < 73.072$

From this, we get,

$45.344 < \log_x 7^{100} < 45.46$

From this we can say, $\lfloor \log_x 7^{100} \rfloor = \boxed{45}$

we obtain values that are slightly larger and slightly smaller than $\log_x 7$ from adding, multiplying and subtracting $\log_x 2$ $\log_x 3$ and $\log_x 5$

$0.441 < \log_x (\frac{20}{3}) < 0.443$

$0.459 < \log_x (\frac{36}{5}) < 0.462$

$0.443 < \log_x 7 < 0.459$ , $log_x 7^{100} = 100 \log_x 7$

$44.3 < 100 \log_x 7 < 45.9$

Therefore, $\left \lfloor \log_x 7^{100} \right \rfloor = 45$

We know from statement above that log x 2 = (0.161 + 0.162)/2 = 0.1615. Then log x 2= log 2/ log x = 0.1615 ---> log x = log 2 / 0.1615 = 0.30130/0.1615 = 1.86563467... We know if log x 7^100 = 100 * (log 7/ log x) = 100* 0,84509804/1.86563467 = 0.4529... x 100 = 45,29.... So | 45.29 _| = 45. Answer : 45

We know from statement above that log x 2 = (0.161 + 0.162)/2 = 0.1615. Then log x 2= log 2/ log x = 0.1615 ---> log x = log 2 / 0.1615 = 0.30130/0.1615 = 1.86563467... We know if log_x 7^100 = 100 * (log 7/ log x) = 100* 0,84509804/1.86563467 = 0.4529... x 100 = 45,29.... So | 45.29 | = 45. Answer : 45

Remember that $7^{4}=2401 = 2^{5} \times 3 \times 5^{2} + 1$ . As the second derivative of natural logarithm is always negative, then (ln(2401) <

I tried the bounds log(4x5/3) and log(4x9/5)

i.e. log(6.67) and log(7.2) which gave 0.441 and 0.459, and I could solve in 2nd attempt.

Dont know why log(48) and log(50) simply didn't strike in the first place :p

logx(7) = log10(7) * logx(10) = 0.8451* (logx(2)+logx(5)) = 0.4508 Multiplying by 100 and rounding to nearest lower integer, we get 45

First I calculated the vale of x by converting the base with log base rule: logx (b) = log10(b)/log10(x) and refined its value for all of the inequalities given in the problem. Finally for a base value x= 76, I continued the problem. Since I have to determine the floor function so i have to just take the largest integer value of 100 log76(7) and which yields me 45!