Asked by Alyssa Pacheco
                Suppose h(x) = 3x and k(x) = log3(x). What is k(h(x))? What about h(k(x))? Start with a few numbers to convince yourself. Explain completely why this is true.
            
            
        Answers
                    Answered by
            Alyssa Pacheco
            
    oops already did that one
i just need to know how to do this other problem.
Since a logarithm is the interchange (or inverse) of an exponential function, each logarithmic function has a particular base. Note that we write the base below the line (like a subscript). For example, we write log2(x). This looks a bit confusing, since the "x " is above the 2. It almost looks like 2x, but it isn't. When you write it, make it very clear. Most log equations can be written as an exponential equation and vice versa, as you saw in the previous problem. Copy each equation shown below. Then rewrite each equation in the other form.
a) y = 5x
b) y = log7(x)
c) 8x = y
d) AK = C
e) K = logA(C)
f) log (k)=N
1/2
    
i just need to know how to do this other problem.
Since a logarithm is the interchange (or inverse) of an exponential function, each logarithmic function has a particular base. Note that we write the base below the line (like a subscript). For example, we write log2(x). This looks a bit confusing, since the "x " is above the 2. It almost looks like 2x, but it isn't. When you write it, make it very clear. Most log equations can be written as an exponential equation and vice versa, as you saw in the previous problem. Copy each equation shown below. Then rewrite each equation in the other form.
a) y = 5x
b) y = log7(x)
c) 8x = y
d) AK = C
e) K = logA(C)
f) log (k)=N
1/2
                    Answered by
            Alyssa Pacheco
            
    so i tried a and i got 
log ^(y)=x
5
Correct?
  
    
log ^(y)=x
5
Correct?
                    Answered by
            Damon
            
    I think you might mean
y = 5^x power
then
log5 (y) = x log5 (5) but log5 (5) = 1
log5 (y) = x
-------------------------
y = log7 (x)
7^y = 7^[log7(x)] but base^[logbase(x)]=x
7^y = x
    
y = 5^x power
then
log5 (y) = x log5 (5) but log5 (5) = 1
log5 (y) = x
-------------------------
y = log7 (x)
7^y = 7^[log7(x)] but base^[logbase(x)]=x
7^y = x
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