alogb
It means that b is based on the logarithm of a.
manufacture
alogb^c
=
x
c*alogb
=
y
∴
a^x
=
b^c
It's also VIII
a^(y/c)=b
a^y
=
b^c
∴a^x
=
a^y
∴
x
=
y
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Making alogb
=
x
alogc
=
y
alog(bc)=z
therefore
b
=
a^x
c
=
a^y
B.C.
=
a^z
Here we go again: b*c
=
a^x
*
a^y
therefore
b*c
=
a^(x+y)
∴
x+y
=
z
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Subtraction can also be proved in the same way.
2~ let f(a)=x
∴
f(x)=3
Solve by the problem
f(x)
=
{x^2
-
4x
+
six
x & gt=0
f(x)=3
therefore
x
={
three
{-x^2
+
four
x & lt0
{- 1
∫f(a)
=
x
The same understanding is possible.
When x=3
When x=- 1
a={3
A={ no solution
{- 1
{-√5
So a
=
{3
{- 1
{-√5
Note: invalid solution should be removed according to the range of X.
3~ attention
For index
a^x
When x= a certain value, no matter what a is, it is always equal to the value of a.
therefore
Can only be 1
a^0
=
1
So (x+ 1) (x 2-5x+6) = 0.
Solve it
x=- 1，2，3
f(- 1，2，3)= 0
eight
therefore
transverse section
(- 1,8)
,
(2,8)
,
(3,8)
4~ ling
After moving, it was
f(y)
∴f(x)
=
Female (y-2)
You ∵f(y)
=
3^y
∴f(y-2)
=
3^(y-2)
∴f(x)=3^(y-2)
5~ same as 4
f(x)
=
f(y)+3
f(y)=3x+4
f(y)+3=3x+7
f(x)
=
3x+7
6 ringy
=
e^z
∵e & gt； 1
∴ e z is increasing monotonically around the world.
You z
=
The symmetry axis of -x 2+4x+ 12 is
-b/2a=2
And a < 0.
Monotonically increasing ∴z on x
Monotonically increasing ∴y on x
7 ~∫f(0)
=
four
∴
manufacture
f(x)
=
ax^2+bx+4
So f(x+2)
=
a(x+2)^2+b(x+2)+4
So f(x+2)-x
=
a(x+2)^2+b(x+2)+4-(ax^2+bx+4)
=
8x+ 14
∴a(x+2-x)(x+2+x)+b(x+2-x)
=
8x+ 14
4ax
+4a+2b
=
8x+ 14
undetermined coefficient
4a
=
eight
a
=
2
4a+2b
=
14
2b
=
six
b
=
three
∴
f(x)
=
2x^2+3x+4
(2)~
∵ the symmetry axis of f (x) is
-b/2a
=
-3/4
∈[- 1, 1]
Shear a>0
The minimum value of f (x) is
(4ac-b^2)/4a
=
23/8
F(x) is greater than y=3x+2m, so the minimum value of f (x) is greater than the maximum value of y=3x+2m.
∫y = 3x+2m monotonically increases.
The maximum value is
3* 1+2m=2m+3
∴2m+3
& lt
23/8
2m & lt- 1/3
m
& lt
- 1/6
See for yourself and correct your mistakes.