What is k>= ((log_{10}(n)))/((log_{10)(2))} ?

The solution to k>= ((log_{10}(n)))/((log_{10)(2))} is log_{10}(n)<= klog_{10}(2)

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k>= ((log_{10}(n)))/((log_{10)(2))}

k \geq \frac{( lo g _{10} ( n ) )}{( lo g _{10} ( 2 ) )}

lo g_{10} (n) \leq k lo g_{10} (2)

Solution steps

k \geq \frac{lo g _{10} ( n )}{lo g _{10} ( 2 )}

Switch sides

\frac{lo g _{10} ( n )}{lo g _{10} ( 2 )} \leq k

Let

u = lo g_{10} (n)

\frac{u}{lo g _{10} ( 2 )} \leq k

\frac{u}{lo g _{10} ( 2 )} \leq k : u \leq k lo g_{10} (2)

u \leq k lo g_{10} (2)

Substitute back

u = lo g_{10} (n)

lo g_{10} (n) \leq k lo g_{10} (2)

lo g_{x + 4} (5 x + 20) \leq lo g_{x + 4} ((x + 4)^{2})

lo g_{x} (x^{2} - 4 x + 4) \geq 0

lo g_{7} (\frac{( 2 x - 6 )}{( 2 x - 1 )}) > 0

(lo g_{0.5} (x))^{2} + lo g_{0.5} (x) - 2 \leq 0

2 - lo g_{10} (4 x) > 0

What is k>= ((log_{10}(n)))/((log_{10)(2))} ?
The solution to k>= ((log_{10}(n)))/((log_{10)(2))} is log_{10}(n)<= klog_{10}(2)