#5(2^(3x-2))=8^(5x+1)# How do I solve for x exactly?

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#5(2^(3x-2))=8^(5x+1)#

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Answer 1 · 2018-03-14T05:03:28

# x=-1/12log_2 6.4#

#5(2^(3x-2))=8^(5x+1)#.

Using the usual rules of indices, we have,

#5{2^(3x)*2^(-2)}=8^(5x)*8^1#,

#:. 5[{(2^3)^x}*1/2^2]={(2^3)^(5x)}{2^3}#,

#:. 5{2^(3x)/2^2}=2^(15x)*2^3#,

#:. 2^(3x)=2^(15x)*2^3*2^2/5#,

#:. 2^(3x)/2^(15x)=2^5/5#,

#:. 2^(3x-15x)=2^5/5#,

#:. 2^(-12x)=32/5=6.4#,

Taking #log_2#, we get,

# log_2(2^(-12x))=log_2 6.4, i.e., #

# -12xlog_2 2=-12x=log_2 6.4, or, x=-1/12log_2 6.4#