How do you calculate ${log}_{2} 4 - {log}_{4} 2$ $log_2 4 - log_4 2$ ?

Question

3109 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-01T09:05:57

${log}_{2} 4 - {log}_{4} 2$ $log_(2)4-log_(4)2$ $=$ $=$ $2 - \frac{1}{2}$ $2-1/2$ $=$ $=$ $\frac{3}{2}$ $3/2$

When I write ${log}_{a} b$ $log_(a)b$ $=$ $=$ $c$ $c$ , in effect I ask to what power do I raise the base $a$ $a$ to get $b$ $b$ .

By this definition, if ${log}_{a} b$ $log_(a)b$ $=$ $=$ $c$ $c$ , then $a^{c} = b$ $a^(c)=b$ .

Typically we use $logs$ $"logs"$ to the base $10$ $10$ or the base $e$ $e$ .

Can you tell me what is ${log}_{10} 100 + {log}_{10} 1000$ $log_(10)100+log_(10)1000$ $?$ $?$

How do you calculate log24−log42log_2 4 - log_4 2?