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,
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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