Can someone solve this equation please? #3^(x+1)=243^(x+2)#

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Answer 1 · 2018-02-22T19:07:27

#x =-9/4 #

This is an exponential equation where the variable is in the index.

The first step s to make the bases the same.

It helps to know that #3^5 = 243#

#3^(x+1) = color(blue)(243)^(x+2)#
#color(white)(wwwww)darr#
#3^(x+1) = (color(blue)(3^5))^(x+2)#

#3^(x+1) = 3^(5x+10)#

If the bases are equal then the indices must be equal.

#:. x+1 = 5x+10#

#1-10 = 5x-x#

#-9 = 4x#

#-9/4 = x#

Answer 2 · 2018-02-22T22:48:30

#x=-2.25 or -9/4#

As an addition to EZ as pi's response, you can use logarithms as a more general approach. Logarithms will work with most exponential equations.

Take the log of both sides:

#log3^(x+1)=log243^(x+2) =>#

#(x+1)log3=(x+2)log243=>#

#(x+1)0.47712=(x+2)2.3856=>#

#0.47712x+0.47712=2.3856x+4.7712#

#-1.90848x=4.29408#

#x=-2.25 or -9/4#