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

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

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

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