How do you solve #4^ { 2x } \cdot 4^ { 3} = 1#?

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Answer 1 · 2017-05-04T20:50:10

#x=-1.5#

#4^(2x)*4^3=1#

law of indices:

#a^m*a^n= a^(m+n)#

using this:

#4^(2x)*4^3 = 4^(2x+3)=1#

#4^0 = 1#

#therefore 2x+3=0#

#2x=-3#

#x=-1.5#

Answer 2 · 2017-05-05T04:22:49

#x = - 1.5#

Firstly, we should probably explore the law of exponents:

= #a^m * a^n = a^(m+n)#

Then, we can apply use this law to calculate #x#

= #4^(2x) * 4^3#
= #4^(2 + 3 )#
= #4^(2x + 3) = 1#

We can also conclude that #4^0 = 1#

#therefore 4^(2x+3) = 4^0#

Now lets use simple algebra to calculated #x#

= #2x + 3 = 0#
= #2x = 0-3#
= #2x = -3#
= #x = -3/2#
= #x = -1.5#