Question

log2 4 * log48 * logg 16
.....nth term = 49, what is the value of n?​

Answer 1

Answer:

137 answer

piyushkumar

Answer 2

We need to recall base change formula of logarithm

$log_{a}b=\frac{log_{c}b }{log_{c}a }$

Given:

   $log_{2}4.log_{4}8.log_{8}16..........=49$

We use base change formula,

$log_{2}4.\frac{log_{2}8 }{log_{2}4}.\frac{log_{2}16 }{log_{2}8 } .....\frac{log_{2}2.2^{n}}{log_{2}2^{n} } } =49$

                             $log_{2 } 2.2^{n} =49$

                    $log_{2 }2+log_{2 }2^{n} =49$

                           $1+nlog_{2} 2=49$

                                   $n+1=49$.......................[$log_{2} 2=1$]

                                         $n=48$