Question: I have to find the tangent lines to the function 
$$f(x)=(x+1)^{3/2}$$
that pass through the point $(4/3, 3)$. I found the derivative of the function 
$$f'(x)=\frac{3\sqrt{x+1}}{2}$$ 
and know that that will be the slope of the two lines. I think I use the point-slope form to get the equation $f(x)-3=f'(x)(x-4/3)$. From here I try solving for $x$ and get absolutely nowhere.

Answer: 
Let's have a look where $a^b$ is well defined in the context of **real numbers**:

* For $a>0$ there is no issue, $a^b$ is defined whatever $b$ is.

In this case we define $a^b=\exp(b\ln(a))$ 

Since $\exp$ and $\ln$ are monotonic thus injective then 

$a^b=a^c\iff (b-c)\ln(a)=0\iff \quad b=c\quad$ or $\quad a=1$

When $a=1$ then $1^b=1^c=1$ for any $b,c$.

* For $a=0$ then $0^b=0$ is defined for $b>0$.

In this case $0^b=0^c$ is true whenever $b>0$ and $c>0$

* For $a<0$ then $a^b$ is defined everywhere $(-1)^b$ is defined, that is $b=\frac pq$ where $(p,q)\in\mathbb Z^2$ and $\gcd(p,q)=1$ and $q$ is odd.

The last condition seems a bit complicated, but remember that $a^{\frac 1q}$ is defined as the solution of $x^q=a$.

Since $x^2\ge 0$ then all even powers of $x$ are positive and $x^q=a$ as no solution when $q$ is even.

The other condition $\gcd(p,q)=1$ ensures not only that the rational $\frac pq$ is reduced to the standard form but also that $p$ and $q$ cannot be both even at the same time to avoid issues like $\sqrt[3]{-1}=(-1)^{\frac 13}=(-1)^{\frac 26}$ and we would be puzzled that $6$ is an even $q$...

In this case $a\neq -1$ and $a^b=a^c\iff b=c$

Finally when $a=-1$ the equation $(-1)^b=(-1)^c$ requires that both $b$ and $c$ are rational with the conditions cited above, but also that their numerator has the same parity.

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To sumarize solving $$h(x)^{f(x)}=h(x)^{g(x)}$$

Let call $\mathcal Q\_{odd}$ the condition: 

$x\in\mathcal Q\_{odd}\iff x=\frac pq$ and $(p,q)\in\mathbb Z\times \mathbb N^\*$ and $\gcd(p,q)=1$ and $q$ is odd.

For $x\in\mathcal Q\_{odd}$ let's call $\mathcal N(x)=p$.

We need to check the following cases:

* $h(x)=0$ and $f(x)>0$ and $g(x)>0$
* $h(x)=1$
* $h(x)>0$ and $f(x)=g(x)$
* $h(x)=-1$ and $f(x)\in\mathcal Q\_{odd}$ and $g(x)\in\mathcal Q\_{odd}$ and $\mathcal N(f(x)),\ \mathcal N(g(x))$ have same parity
* $h(x)<0$ and $f(x)=g(x)\in\mathcal Q\_{odd}$

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Note that in your solution $x=1$ belongs to bullet $\#5$ and not $\#3$ since $h(x)>0$ requires $x>3$. We can verify that exponents $f(1)=g(1)=7$ are integers (thus indeed odd denominator rationals). So $(-2)^7$ is well defined.
Yes, for the exponential equation $f(x)=b^x$ with the restrictions that $b>0$ and $b\ne1$, cases $2,\,3$ and $4$ are eliminated, so only case $1$ applies.

What this means is that if $b>0$ and $b\ne1$ and if $f(x)=b^x$ then it necessarily follows that

$$f(x\_1)=f(x\_2)\text{ if and only if }x\_1=x\_2$$

A function with this property is called a **one-to-one** function.