1/3 ln(<em>x</em>) + ln(2) - ln(3) = 3
Recall that
, so
ln(<em>x</em> ¹ʹ³) + ln(2) - ln(3) = 3
Condense the left side by using sum and difference properties of logarithms:


Then
ln(2/3 <em>x</em> ¹ʹ³) = 3
Take the exponential of both sides; that is, write both sides as powers of the constant <em>e</em>. (I'm using exp(<em>x</em>) = <em>e</em> ˣ so I can write it all in one line.)
exp(ln(2/3 <em>x</em> ¹ʹ³)) = exp(3)
Now exp(ln(<em>x</em>)) = <em>x </em>for all <em>x</em>, so this simplifies to
2/3 <em>x</em> ¹ʹ³ = exp(3)
Now solve for <em>x</em>. Multiply both sides by 3/2 :
3/2 × 2/3 <em>x</em> ¹ʹ³ = 3/2 exp(3)
<em>x</em> ¹ʹ³ = 3/2 exp(3)
Raise both sides to the power of 3:
(<em>x</em> ¹ʹ³)³ = (3/2 exp(3))³
<em>x</em> = 3³/2³ exp(3×3)
<em>x</em> = 27/8 exp(9)
which is the same as
<em>x</em> = 27/8 <em>e</em> ⁹
If you can find one leg of a triangle to be congruent to a leg on the other triangle, then you can use the HL (hypotenuse leg) theorem. If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and a leg of another right triangle, then the triangles are congruent.
Answer:
![f^{-1}(x)=\sqrt[3]{x}-6](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D-6)
Step-by-step explanation:



![\sqrt[3]{x}=y+6](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%3Dy%2B6)
![\sqrt[3]{x}-6=y](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D-6%3Dy)
The answer is B, C, and D. Like terms are terms with all the same variable, so 5x and -x are like terms.
C is correct. If we add -x to 5x, we get 4x. The other numbers remain unchanged because they have no like terms.
D is correct. Applying the rule of like terms, which is that like terms are numbers with the same variable, only add together numbers with the same variable.
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