MHB How to Solve Exponential Equations?

[Alexeia]

Hi,

I attached a pdf with a question and its answers. I don't understand the whole thing basically, e.g what their asking for..

1.2.1;1.2.2;1.2.3

1.2.2 - How many solutions for x will the equation have..?
1.2.3 - Largest integer for which it will have solutions..?

Any explanations will help..

Thanks

 

[MarkFL]

Here are the questions:

1.2 Given: $$2^x+2^{x+2}=-5y+20$$

1.2.1 Express $$2^x$$ in terms of $y$.

Here, you are expected to solve the equation for $2^x$. Recall the property of exponents:

$$a^{b+c}=a^ba^c$$

which means:

$$2^{x+2}=2^x2^2=4\cdot2^x$$

So now the equation becomes:

$$2^x+4\cdot2^x=-5y+20$$

Can you continue? Combine the two like terms on the left side as your first step.

After we get this part solved, where you understand it, we can move on to the remaining parts, as they depend on this part.

 

[Alexeia]

Here are the questions:

1.2 Given: $$2^x+2^{x+2}=-5y+20$$

1.2.1 Express $$2^x$$ in terms of $y$.

Here, you are expected to solve the equation for $2^x$. Recall the property of exponents:

$$a^{b+c}=a^ba^c$$

which means:

$$2^{x+2}=2^x2^2=4\cdot2^x$$

So now the equation becomes:

$$2^x+4\cdot2^x=-5y+20$$

Can you continue? Combine the two like terms on the left side as your first step.

After we get this part solved, where you understand it, we can move on to the remaining parts, as they depend on this part.

Awesome, the way the questioned was asked confused me.. Another way of saying it then is - Solve for 2^x.. But it doesn't yet mean to solve for x right?

Then Question 1.2.3, can you shed some light on that one pls?

 

[Evgeny.Makarov]

Then Question 1.2.3, can you shed some light on that one pls?

1.2 Given:
\[
2^x+2^{x+2}=-5y+20\tag{1}
\]
...
1.2.3: Solve for $x$ if $y$ is the largest possible integer value for which (1) will have solutions.

Question 1.2.3 (admittedly, a bit convoluted) says the following. Let $P(y)$ be some property of $y$ (to be discussed later). Find the largest integer $y$ such that $P(y)$ holds and call it $y_{\text{max}}$. Now solve (1) for $x$ assuming that $y=y_{\text{max}}$.

The property $P(y)$ in question is that (1) has a solution $x$ for a given $y$. Note that the equation
\[
2^x+2^{x+2}=z
\]
has some solution $x$ iff $z>0$. Therefore,
\[
P(y)\text{ holds }\iff -5y+20>0 \iff y<4.
\]
The largest $y_{\text{max}}$ satisfying this property is 3 (since the inequality is strict). Therefore, you are supposed to solve (1) when $y=3$.