Solving equations using substitution

Exponential functions and logarithms: Logarithmic functions

Solving equations using substitution

Previously, we solved complicated equations using substitution. We can also use this for logarithms and exponential equations. When we write #\log_a\!\left(x\right)^2#, we mean #\left(\log_a\left(x\right)\right)^2#. We'll give a step-by-step guide for solving these equations. For exponential functions we also need to check whether or not the solutions are valid.

Solving logarithmic equations using substitution

	Step-by-step	Example
	We solve a logarithmic equation with unknown #x# using substitution and the quadratic formula.	#\log_2\!\left(x\right)^2+3\cdot\log_2\left(x\right)=4#
Step 1	Choose a suitable substitution #\blue{u}=\log_\green{a}\left(x\right)#.	#\blue{u}=\log_\green{2}\left(x\right)#
Step 2	Substitute.	#\blue{u}^2+3\blue{u}=4#
Step 3	Move everything to the left, so that we can apply the quadratic formula.	#\blue{u}^2+3\blue{u}-4=0#
Step 4	Apply the quadratic formula.	#\blue{u}=-4\vee\blue{u}=1#
Step 5	Now solve the equation for #x#	#\log_\green{2}\left(x\right)=-4# gives #x=\frac{1}{16}# #\log_\green{2}\left(x\right)=1# gives #x=2#

Solving exponential equations using substitution.

	Step-by-step	Example
	We solve a exponential equation with unknown #x# using substitution and the quadratic formula	#9^x=6+3^x#
Step 1	Write the exponents with equal base.	#(3^x)^2=6+3^x#
Step 2	Choose a suitable substitution #\blue{u}=\green{a}^x#.	#\blue{u}=\green{3}^x#
Step 3	Substitute.	#\blue{u}^2=6+\blue{u}#
Step 4	Move everything to the left, so that we can apply the quadratic formula.	#\blue{u}^2-\blue{u}-6=0#
Step 5	Apply the quadratic formula.	#\blue{u}=3 \vee \blue{u}=-2#
Step 6	Now solve the equation for #x#, if possible.	#\green{3}^x=3# gives #x=\log_{\green{3}}\left(3\right)=1# #\green{3}^x=-2# gives #x=\log_{\green{3}}\left(-2\right)#, no solution.

Solve the following equation for #x#, by means of substitution.
\[\log_3\left(x\right)^2-4\cdot\log_3\left(x\right)=-3\]

Write:

#none# if there is no solution,
#x=x_1# if there is one solution,
#x=x_1\lor x=x_2# if there are two solutions,

with the correct exact values for #x_1# and #x_2#.

#x=3 \vee x=27#

We'll use the step-by-step guide

Step 1	Choose a suitable substitution #u=\log_3\left(x\right)#.
Step 2	Substitute, which gives the equation \[u^2-4u=-3\]
Step 3	Move everything to the left, so that we can apply the quadratic formula \[u^2-4u+3=0\]
Step 4	Apply the quadratic formula, which gives us the following solutions for #u#. \[u=1 \vee u=3\]
Step 5	Now solve the equation for #x#. #\log_3\left(x\right)=1# gives #x=3# and #\log_3\left(x\right)=3# gives #x=27#.

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