MHB Evaluate logarithm of a number

[anemone]

Given that

$0.375<\log_x 5<0.376$

$0.256<\log_x 3<0.257$

$0.161<\log_x 2<0.162$

evaluate $\lfloor \log_x 7^{100} \rfloor$.

 

[eddybob123]

Spoiler

By brute force, we see that $x=73$ satisfy all of the inequalities, hence,
$$\lfloor\log_x{7^{100}}\rfloor$$
$$=\lfloor100\log_{73}{7}\rfloor$$
$$=45$$

 

[Albert1]

Given that

$0.375<\log_x 5<0.376$

$0.256<\log_x 3<0.257$

$0.161<\log_x 2<0.162$

evaluate $\lfloor \log_x 7^{100} \rfloor$.

Spoiler

$0.470<\log_x 7.5=\log_x \dfrac {3\times 5}{2}=\log_x 3+\log_x 5 -\log_x 2 <0.471$
now I will use linear interpolation :
let:$y=\log_x 7$
$\dfrac {\log_x {7.5 -y}}{7.5-7}\approx \dfrac {\log_x {7.5 -\log _x 5}}{7.5-5}$
$ y \approx 0.453$
$\therefore \lfloor \log_x 7^{100} \rfloor=45$

 

[Albert1]

I t seemed not easy to get $\log_x 7,\, directly \,\, from ,\log _x 2, \log_x 3, and \log _x 5$
instead I get $\log_x 7.5$
so I use the method of linear interpolation

 

[Opalg]

Given that

$0.375<\log_x 5<0.376$

$0.256<\log_x 3<0.257$

$0.161<\log_x 2<0.162$

evaluate $\lfloor \log_x 7^{100} \rfloor$.

[sp]$\log_x 48 = 4\log_x2 + \log_x3 >4\times0.161 + 0.256 = 0.9$.

$\log_x50 = \log_x2 + 2\log_x5 < 0.162 + 2\times 0.376 = 0.914$.

Therefore $0.9 < \log_x49 < 0.914$. But $49 = 7^2$, so $\log_x7 = \frac12\log_x49$ and $0.45 < \log_x7 < 0.457$. Finally, $45 < \log_x 7^{100} < 45.9$ and so $\lfloor \log_x 7^{100} \rfloor = 45$.[/sp]

 

[anemone]

Thank you all for participating and yes, 45 is the correct answer!:)