#1) Inverse Functions, note log(x) in SAGE is ln(x)
#1) f(g(x))==g(f(x)) iff g(x) is the inverse of f(x)
#1) f(x) is a reflection of g(x) about y=x iff g(x) is the inverse of f(x)
f(x)=e^x
g(x)=log(x)
show(f(g(x)))
show(g(f(x)).simplify())
show(bool(f(g(x))==g(f(x)).simplify()))
plot([f(x),g(x),x],0,2,aspect_ratio=1)
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#2) graph y=-ln(x+2) using what you know about the graph of y=ln(x)
f(x)=ln(x)
g(x)=-ln(x)
h(x)=-ln(x+2)
plot([f(x),g(x)],0,2,aspect_ratio=1)
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plot([h(x)],-2,0,aspect_ratio=1)
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#3) solve for x: log base 3 (4x-7)==2
solve([log(4*x-7)/log(3)==2],x)
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plot([log(4*x-7)/log(3),2],2,5,aspect_ratio=1)
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#4) solve for x: log base x 64 == 2
ans=solve([log(64)/log(x)==2],x);show(ans)
show(ans[0])
show(ans[0].rhs())
show(ans[0].rhs().simplify_radical())
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plot([log(64)/log(x),2],6,9,aspect_ratio=1)
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#5) solve for x: e^(2*x)==5
ans=solve([e^(2*x)==5],x);show(ans)
show(ans[0].rhs().n())
show(ans[1].rhs().n())
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plot([e^(2*x),5],0,1)
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Note 5a) using ln:
(let ln means log base e)
e^(2*x)==5
ln(e^(2*x))==ln(5)
2*x==ln(5)
x==ln(5)/2
Note 5b)using log:
(let log means log base 10)
e^(2*x)==5
log(e^(2*x))==log(5)
2*x*log(e)==log(5)
x==1/2*log(5)/log(e)
Note 5c) change of base formula (from unknown base to known base logs)!
therefore, log base b of a = (known log of a)/(known log of b)
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