How do you solve #a^x = 10^(2x+1)#?

Answer 1

#x=(-log(10))/(2log(10)-log(a))#

#1#. Assuming you are trying to solve for #x#, start by taking the logarithm of both sides.
#a^x=10^(2x+1)#
#log(a^x)=log(10^(2x+1))#
#2#. Using the logarithmic property, #log_color(purple)b(color(red)m^color(blue)n)=color(blue)n*log_color(purple)b(color(red)m)#, simplify the equation.
#xlog(a)=(2x+1)log(10)#
#3#. Expand the brackets.
#xlog(a)=2xlog(10)+log(10)#
#4#. Move all terms with #x# to one side of the equation with the terms with no #x# to the other side.
#2xlog(10)-xlog(a)=-log(10)#
#5#. Factor out #x#.
#x(2log(10)-log(a))=-log(10)#
#6#. Isolate for #x#.
#color(green)(|bar(ul(color(white)(a/a)x=(-log(10))/(2log(10)-log(a))color(white)(a/a)|)))#
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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