How do you solve #7^(x+2)=e^(17x)#?

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Answer 1 · 2016-01-10T09:16:05

#x=(-2ln(7))/(ln(7)-17)#

Using the following principle
If you have log to base b of b#-> log_b(b)=1#

Given: #7^(x+2)=e^(17x)#

Taking logs

#ln(7^(x+2))=ln(e^(17x))#

#=> (x+2)ln(7)=17xln(e)#

But #ln(e)=1 color(white)(.)#giving:

#(x+2)ln(7)=17x#

Multiply out the bracket

#xln(7)+2ln(7)=17x#

Collecting like terms

#xln(7)-17x=-2ln(7)#

#x=(-2ln(7))/(ln(7)-17)#