SOLUTION: Please help me solve this problem: log16x+log4x+log2x=7. I know you have to use the change of base theorem to get: logx/log16+logx/log4+logx/log2=7, but I'm not sure what to do aft
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-> SOLUTION: Please help me solve this problem: log16x+log4x+log2x=7. I know you have to use the change of base theorem to get: logx/log16+logx/log4+logx/log2=7, but I'm not sure what to do aft
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Question 708362: Please help me solve this problem: log16x+log4x+log2x=7. I know you have to use the change of base theorem to get: logx/log16+logx/log4+logx/log2=7, but I'm not sure what to do after that. Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! Please help me solve this problem:
log16x+log4x+log2x=7
convert to base 2:
log2(x)/log2(16)+log2(x)/log2(4)+log2(x)=7
log2(x)/4+log2(x)/2+log2(x)=7
LCD:4
log2(x)+2log2(x)+4log2(x)=4*7
log2[x*(x)^2*x^4]=28
log2[x*x^2*x^4]=28
log2[x^7]=28
convert to exponential form
2^28=x^7
x^7=2^28
x=(2^28)(1/7)
x=2^4
x=16
check: Using log key on calculator
log16(x)=log(16)/log(16)=1
log4(x)=log(16)/log(4)=2
log2(x)=log(16)/log(2)=4
log16x+log4x+log2x=1+2+4=7
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