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 stepbystep 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
Stepbystep  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.
Stepbystep  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.
\[\log_3\left(x\right)^2+1\cdot\log_3\left(x\right)=6\]
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,
We'll use the stepbystep guide
Step 1  Choose a suitable substitution #u=\log_3\left(x\right)#. 
Step 2  Substitute, which gives the equation \[u^2+u=6\] 
Step 3  Move everything to the left, so that we can apply the quadratic formula \[u^2+u6=0\] 
Step 4  Apply the quadratic formula, which gives us the following solutions for #u#. \[u=2 \vee u=3\] 
Step 5  Now solve the equation for #x#. #\log_3\left(x\right)=2# gives #x=9# and #\log_3\left(x\right)=3# gives #x={{1}\over{27}}#. 
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