Question 166638
I'll do the first two to get you started.



A)


{{{27^(x)=3^(x+1)}}} Start with the given equation



{{{(3^3)^(x)=3^(x+1)}}} Rewrite {{{27}}} as {{{3^3}}}



{{{3^(3x)=3^(x+1)}}} Multiply the exponents.



Since the bases are equal, this means that the exponents are equal



{{{3x=x+1}}} Set the exponents equal to one another.



{{{3x-x=1}}} Subtract {{{x}}} from both sides.



{{{2x=1}}} Combine like terms on the left side.



{{{x=(1)/(2)}}} Divide both sides by {{{2}}} to isolate {{{x}}}.



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Answer:


So the answer is {{{x=1/2}}} which in decimal form is {{{x=0.5}}}. 




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B)


{{{e^(3x-1)=12}}} Start with the given equation



{{{3x-1=ln(12)}}} Take the natural log of both sides to eliminate the base "e"



{{{3x=ln(12)+1}}} Add 1 to both sides.



{{{x=(ln(12)+1)/3}}} Divide both sides by 3 to isolate "x"




So the answer is {{{x=(ln(12)+1)/3}}} which approximates to 1.16164