How do you solve for the power of #x#? For example, #2^x = 423#. How do you get #x#?

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Answer 1 · 2014-10-02T00:23:47

#2^x=423#

Take the natural log of both sides

#ln (2^x)= ln (423)#

Use one of properties of logs to move the exponent down as a factor

#x*ln (2)=ln(423)#

Use Algebra to solve for #x# by dividing by #ln(x)#

#(x*ln (2))/(ln(2))=ln(423)/(ln(2))#

Use a calculator to resolve the division

#x=ln(423)/(ln(2))=8.724513853#