How do you solve the equation #0.16^(4+3x)=0.3^(8-x)#?

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Answer 1 · 2017-01-29T21:51:22

#x=0.29#

#0.16^(4+3x)=0.3^(8-x)#

Take the log of both sides

#log0.16^(4+3x)=log0.3^(8-x)#

#loga^b=bloga#

#(4+3x)log0.16=(8-x)log0.3#

#4+3x=log0.3/log0.16(8-x)#

Note: from this point on, it is better to use the exact value of #log0.3/log0.16#, but for the sake of simplicity, a rounded value was used.

#4+3x=5.33-0.66x#

#4.66x=1.33#

#x=1.33/4.66=0.29#