How do you solve #3^(4x) = 3^(5-x)#?

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Answer 1 · 2016-05-02T16:49:34

#x=1#

#color(blue)("Quickest way using shortcuts")#
#color(brown)("If you have "x^a=x^b" then "a=b)#

#=>4x=5-x#

#4x+x=5#

#5x=5 => x=1#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Another method that demonstrates other properties of indices.") #

Note that #3^(5-x)" is the same as "3^5/(3^x#

Write as #3^(4x)=3^5/3^x#

Multiply both sides by #3^x#

#3^(4x)xx3^x=3^5 xx3^x/3^x#

But #3^x/3^x=1" and "3^(4x)xx3^x=3^(5x)# giving

#3^(5x)=3^5#

Comparing just the indices

#5x=5#

#=>x=1#

Answer 2 · 2016-05-02T21:52:57

Notice from the outset that the two exponents must be equal, since their bases are both #3#.

#color(red)3^color(blue)(4x)=color(red)3^color(blue)(5-x)" "=>" "color(blue)(4x)=color(blue)(5-x)" "=>" "5x=5#

Thus, #x=1#.

Answer 3 · 2016-05-02T22:50:17

#x=1#

As a Real-valued function of Real numbers, #f(x) = 3^x# is strictly monotonically increasing and is therefore one-to-one.

So #f(4x) = 3^(4x) = 3^(5-x) = f(5-x)# implies #4x = 5-x#

Add #x# to both sides to get:

#5x=5#

Divide both sides by #5# to get:

#x=1#