How do you solve #2^x=4.5#?

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Answer 1 · 2016-07-22T07:43:34

#x=(ln4.5)/(ln2) ~~ 2.17#

Take logs of both sides:

#ln2^x = ln4.5#

Using power rule of logs, bring the x down in front

#xln2 = ln4.5#

Rearrange:

#x=(ln4.5)/(ln2)#

Answer 2 · 2016-07-22T07:43:59

#x=log 9/log 2-1=2.16992500144#

Given #2^x=4.5#, Find #x#

Solution:

#2^x=4.5#

Take the logarithm of both sides of the equation

#log 2^x=log 9/2#

#x*log 2=log 9-log 2#

divide both sides of the equation by #log 2#

#(x*log 2)/log 2=(log 9-log 2)/log 2#

#(x*cancellog 2)/cancellog 2=(log 9-log 2)/log 2#

#x=log 9/log 2-1#

#x=2.16992500144#

God bless....I hope the explanation is useful.