There's Too Many Logarithms To Calculate

Relevant wiki: Properties of Logarithms - Basic

$\large \log _{ 2 }{ 3 } \times \log _{ 3 }{ 4 } \times \log _{ 4 }{ 5 } ...\log _{ 127 }{ 128 }$

$\large= \frac { \log { 3 } }{ \log { 2 } } \times \frac { \log { 4 } }{ \log { 3 } } \times \frac { \log { 5 } }{ \log { 4 } } \times\cdots \times \frac { \log { 128 } }{ \log { 127 } }$

$\large=\frac { \log { 3\quad \times \quad \log { 4 } \quad \times \quad \log { 5 } \quad \times \quad ...\quad \log { 128 } } }{ \log { 2 } \quad \times \quad \log { 3 } \quad \times \quad \log { 4 } \quad \times \quad ...\quad \log { 127 } }$

$\large= \frac { \log { 128 } }{ \log { 2 } }$

$\large= \log _{ 2 }{ 128 }$

$\large =\log _{ 2 }{ { 2 }^{ 7 } }$

$\large=7 \times \log _{ 2 }{ { 2 } }$

$\large=7 \times1$

$\large= \boxed{7}$

$Call:~\log_{2}{3}=a_{1};\log_{3}{4}=a_{2};\cdot\cdot\cdot;\log_{127}{128}=a_{126}$ $\implies 2^{a_{1}}=3;3^{a_{2}}=4;\cdot\cdot\cdot;127^{a_{126}}=128$ $\iff ((\cdot\cdot\cdot((2^{a_{1}})^{a_{2}})^{\cdot\cdot\cdot})^{a_{126}})=128$ $\iff 2^{a_{1} \times a_{2} \times \cdot\cdot\cdot \times a_{126}}=128=2^{7}$ $\iff a_{1} \times a_{2} \times \cdot\cdot\cdot \times a_{126}=7$ Hence, the answer is $\LARGE \boxed{7}$

$\log_2{3}\cdot\log_3{4}\cdot\log_4{5}...\log_{127}{128}$

us the change of base formula we see that all cancel except two values resulting in:

$\frac{\log{128}}{\log{2}}$ = $\log_2{128}$ = $\log_2{2^{7}}$ =7

I will leave it to you to fill in the details as this will help you learn.