How do you solve #2^(x-3) = 5 ^ (x+7)#?

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Answer 1 · 2016-05-04T16:03:48

#=>x~~-14.5647" to 4 decimal places")#

Taking logs

#(x-3)ln(2)=(x+7)ln(5)#

#=>x-3=(x+7)ln(5)/ln(2)#

Let #ln(5)/ln(2)=A#

#x-3=(x+7)A#

#=>x-Ax=7A+3 larr# Corrected from -3 Tony B

#=>x=(7A+3)/(1-A)#

#cancel(=>x~~-10.0259" to 4 decimal places")#

#=>x~~-14.5647" to 4 decimal places")#