A simple problem!

log64=log2^6=6log2=6*0.301=1.806

$\log2=0.301$ (given) $\therefore \, \log \ 64 =\log\ 2^6 =6 \times \log 2 =6 \times 0.301 =1.806$

log64=log2^6=6log2=6*0.301=1.806

log(2)=.301,

So 10^.301=2.

We want to solve 10^x=64 using this information.

We can rewrite 64 as 2^6, so 10^x=(2)^6.

We then rewrite 2 in the equation using the given log, and we get

10^x=(10^.301)^6=10^(6*.301)=10^(1.806).

x=1.806

Therefore log(64)=1.806

log(64)=log(2^6)=6 log(2)=6 0.301=1.806

log64=log2^6=6log2=6*0.301=1.806

$\log { (64) }$

$\log { ({ 2 }^{ 6 }) }$

$6\log { (2) }$

we know that $\log { (2) } =0.301$ so

$6\quad (0.301)$

$1.806$

log64 = log2^6 = 6log2 = 6*0.301 = 1.806

log2=0.301

log64=log2^6

log64=6log2, i.e.

6*0.301= 1.806

gotta use log rules here... 64 is the same as 2 ^6 so the log of 64 is the same as log (2^6) > 6 log (2) and we already know log (2) so just multiply that by 6 and voila!

log64=log2^6=6log2=6*0.301=1.806

log64=log2^6=6*0.301=1.806

step I. express 64 as 2^6; step II. plug in the equation as log 64= log 2^6; step III. from logarithmic laws, we know that log a^b = b log a, therefore the equation becomes 6 log 2 = 6*(0.301) = 1.086

64=2^6. Log 64=6 X Log2 = 6 X 0.301 = 1.806 .Whatever be the BASE.

log(2)=0.301 then log (64)= log(2^6) then by the law of log wherein log(a^b)= blog(a) log(2^6)=6log(20) 60.301=1.806

64=2^6. so, log64 =log2^6 =6log2 =6*.301 =1.806.

log(2)=0.301.. log(64)=log(2^6)=6 log(2).. 6 (0.301)=1.806

There are several ways to do this.

The easiest way is to see that $\log { (64) } =\log { ({ 2 }^{ 6 }) } =6\log { (2) }$ . Then $6\log { (2) } =6\cdot 0.301=1.806$ .

2log64=6log2=6*0.301=1.806

log(2)=0.301 then log (64)= log(2^6) then by the law of log wherein log(a^b)= blog(a) log(2^6)=6 log(20) 6 0.301=1.806