How do you solve #243=9^(2x+1)#?

2 Answers
EZ as pi
Sep 23, 2016

#x = 3/4#

Explanation:

It really is an advantage to know all the powers up to 1000.
You will then recognise that # 243 = 3^5#

Without knowing this, you will have to resort to finding the prime factors. Another clue is that #9 = 3^2#

So this is an exponential equation with 3 as the base,

#9^(2x+1) = 234" "larr# change to the common base of 3

#3^(2(2x+1)) = 3^5" "larr# multiply the indices

#3^(4x+2) = 3^5#

The bases are equal, so the indices must be equal.

#4x+2 = 5#

#4x =3#

#x = 3/4#

A08
Sep 23, 2016

#x=3/4#

Explanation:

#243=9^(2x+1)#
On inspection we see that LHS, #243# can be expressed in terms of powers of #3#,
#243=3^5#
Similarly on RHS, #9# can also be expressed as #(3^2)#
Thus the equation becomes
#3^5=(3^2)^(2x+1)#
#=>3^5=3^(2xx(2x+1))#
#=>3^5=3^(4x+2)#
Since bases on both sides are equal, therefore for the equation to be true, powers must be equal.
#=>5=(4x+2)#
Solving for #x# we get
#x=3/4#

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