How do you solve #7^x + 2 = 410#?

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Answer 1 · 2016-05-20T09:27:34

3.09

#7^x + 2 =410#

#7^x = 408#

Take the natural log of both sides,we get,

#ln(7^x) = ln (408)#

Using logarithm properties, we can bring the x exponent outside the ln

#xln(7) = ln (408)#

#x=ln(408) / ln(7)#

#=(6.0112)#/#(1.945) #

#3.09#