Exponential functions and logarithms: Logarithmic functions
Solving equations using substitution
How do you solve the following equation?
\[4^x=5\cdot 2^x+6\]
Drag the steps so they appear in the correct order.
- Step 1
- Step 2
- Step 3
- Step 4
- Step 5
- Step 6
- Solve the equation for #x#, if possible
- Write the exponents with equal base
- Apply the quadratic formula
- Choose a suitable substitution
- Substitute
- Rewrite it to a form for which we can apply the quadratic formula
- #u=-1# gives #x=\log_2\left( -1\right) #, so is not valid. #u=6# gives #x=\log_2\left( 6\right) #, so is valid
- #u^2=5\cdot u+6#
- #\left( 2^x \right) ^2 =5 \cdot 2^x+6#
- #u=-1 \vee u=6#
- #u^2-5\cdot u - 6=0#
- #u=2^x#
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