Find the equation of the straight line, tangent to $y = e ^ { 7 x } \log _ { 5 } x$ at the point $( 1,0 )$ .
A) $y ( x ) = \frac { e ^ { 7 } } { \ln 5 } x + \frac { e ^ { 7 } } { \ln 5 }$
B) $y ( x ) = \frac { e ^ { 5 } } { \ln 7 } x - \frac { e ^ { 5 } } { \ln 7 }$
C) $y ( x ) = \frac { e ^ { 5 } } { \ln 7 } x + \frac { e ^ { 5 } } { \ln 7 }$
D) $y ( x ) = \frac { e ^ { 7 } } { \ln 5 } x - \frac { e ^ { 7 } } { \ln 5 }$
E) none of these

Question

Find the equation of the straight line, tangent to $y = e ^ { 7 x } \log _ { 5 } x$ at the point $( 1,0 )$ .  
A)  $y ( x ) = \frac { e ^ { 7 } } { \ln 5 } x + \frac { e ^ { 7 } } { \ln 5 }$
B)   $y ( x ) = \frac { e ^ { 5 } } { \ln 7 } x - \frac { e ^ { 5 } } { \ln 7 }$
C)   $y ( x ) = \frac { e ^ { 5 } } { \ln 7 } x + \frac { e ^ { 5 } } { \ln 7 }$
D)   $y ( x ) = \frac { e ^ { 7 } } { \ln 5 } x - \frac { e ^ { 7 } } { \ln 5 }$
E) none of these

Quizplus · Accepted Answer

The derivative of $y = e^{7x} \log_5 x$ at $x = 1$ gives the slope of the tangent line. The derivative is calculated using the product rule and chain rule. At $x = 1$, the slope is $\frac{e^7}{\ln 5}$. The equation of the tangent line is $y = \frac{e^7}{\ln 5}(x - 1)$, which simplifies to $y = \frac{e^7}{\ln 5}x - \frac{e^7}{\ln 5}$.

Find the Equation of the Straight Line, Tangent To y=e7xlog⁡5xy = e ^ { 7 x } \log _ { 5 } xy=e7xlog5​x

Find the Equation of the Straight Line, Tangent To $y = e ^ { 7 x } \log _ { 5 } x$