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:)