How do you solve #2^ { m + 1} + 9= 44#?

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Answer 1 · 2018-03-03T14:29:13

#m=log_2(35)-1~~4.13#

We start by subtracting #9# from both sides:
#2^(m+1)+cancel(9-9)=44-9#

#2^(m+1)=35#

Take #log_2# on both sides:
#cancel(log_2)(cancel(2)^(m+1))=log_2(35)#

#m+1=log_2(35)#

Subtract #1# on both sides:
#m+cancel(1-1)=log_2(35)-1#

#m=log_2(35)-1~~4.13#

Answer 2 · 2018-03-03T14:31:26

#m~~4.129# (4sf)

#2^(m+1)+9=44#

#2^(m+1)=35#

In logarithm form, this is:

#log_2(35)=m+1#

I remember this almost as keep 2 as the base and switch the other numbers.

#m=log_2(35)-1#

#m~~4.129# (4sf)

Answer 3 · 2018-03-03T14:41:09

#m=(log35-log2)/log2#

#2^(m+1)+9=44#

#2^(m+1)=44-9=35#

#log (2^(m+1))=log35" "# (taking the logarithm base #10# on both sides)

#log(2^m * 2)=log35#

#log2^m +log2=log35#

#log2^m=log35-log2#

#mlog2=log35-log2#

#m=(log35-log2)/log2#