How do you solve $10= 2e ^ { 5x }$ ?

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Answer 1 · 2017-10-20T08:59:25

Let's see.

Given $rarr$

$2e^(5x)=10$

Now taking $"log"$ to the base $10$ on both sides:

$log(2e^(5x))=log10$

Now, $log(ab)=loga+logb$ $rarr$

$:.log2+loge^(5x)=log10$

Now, $log(a^b)=bloga$ $rarr$

$:.5xcdotloge+log2=log10$

Value of $e=2.7$ $rarr$

$:.5xcdotlog(2.7)+log2=log10$

$:.(5x)xx(0.43)+0.3=1$

$:.2.15x=1$

$:.x=0.465$

Therefore $color(red)(x=0.465)$ . (Answer).

Hope it Helps:)

How do you solve 10= 2e ^ { 5x }10= 2e ^ { 5x }?