The equation provided is:
5log5(x)+logx(5)=9
Let's solve this step-by-step:
Step 1: Simplify log5(x)
Using the logarithmic property logb(a)=log(b)log(a), we write:
log5(x)=log(5)log(x).
Step 2: Simplify logx(5)
Similarly, for logx(5), we write:
logx(5)=log(x)log(5).
Step 3: Substitute these into the equation
Substituting these into the original equation:
5⋅log(5)log(x)+log(x)log(5)=9.
Let y=log(x), so the equation becomes:
5⋅log(5)y+ylog(5)=9.
Step 4: Eliminate the fraction
Multiply through by y⋅log(5) to eliminate the denominators:
5y2+(log(5))2=9y⋅log(5).
Step 5: Rearrange into a standard quadratic form
Reorganize the terms:
5y2−9y⋅log(5)+(log(5))2=0.
This is a quadratic equation in y, where:
- a=5,
- b=−9⋅log(5),
- c=(log(5))2.
Step 6: Solve using the quadratic formula
The quadratic formula is:
y=2a−b±b2−4ac.
Substituting a, b, and c:
y=2⋅59⋅log(5)±(9⋅log(5))2−4⋅5⋅(log(5))2.
Simplify:
y=109⋅log(5)±81⋅(log(5))2−20⋅(log(5))2.
y=109⋅log(5)±61⋅(log(5))2.
y=109⋅log(5)±log(5)⋅61.
Factor out log(5):
y=log(5)⋅109±61.
Step 7: Back-substitute for y=log(x)
log(x)=log(5)⋅109±61.
Exponentiate both sides to solve for x:
x=5109±61.
Thus, there are two solutions:
x=5109+61andx=5109−61.
Final Answer:
The solutions are:
x=5109+61,x=5109−61.
Would you like a deeper explanation of any step?
Here are 5 related questions to explore:
- How can the quadratic formula be used in logarithmic equations more generally?
- What are the properties of logarithms and how are they applied in problem-solving?
- How do changes of base for logarithms affect their use in equations?
- What happens if logx(5) is written directly as its inverse (1/log5(x))?
- Can these solutions for x be expressed in terms of decimals instead of logarithmic expressions?
Tip: Always check if logarithmic solutions satisfy the domain of the original equation (e.g., x>0,x=1).