# Solve $4^{9x-4} = 3^{9x-4}$

I am having some trouble trying to solve

$$4^{9x-4} = 3^{9x-4}$$

I tried to make each the same base but then I'm becoming confused as to what to do next.

These are the steps I took:

\begin{align} 4^{9x-4} &= 3^{9x-4} \\ \log_4(4^{9x-4}) &= \log_4(3^{9x-4}) \\ \end{align}

Where do I go from there?

Thanks!

• $9x-4=0 \Rightarrow x=\frac{4}{9}$ – Ömer Oct 10 '13 at 12:26
• Can you solve $4^Q=3^Q$? – Gerry Myerson Oct 10 '13 at 12:27
• @GerryMyerson I know how to solve $4^Q = 3$, does the same concept apply to $4^Q = 3^Q$? – Jeel Shah Oct 10 '13 at 12:29
• That would depend on the concept. How do you solve $4^Q=3$? – Gerry Myerson Oct 10 '13 at 12:30
• I would $\log_4 (4^Q) = log_4(3)$. Which would then be Q = $\log_4(3)$ or $\log(3)/\log(4)$ – Jeel Shah Oct 10 '13 at 12:36

Magic property of logarithms: $$\log_b(x^y) = y\log_b(x)$$ So here, $$\log_4(4^{9x - 4}) = (9x-4)\log_4(4)$$ and $$\log_4(3^{9x-4}) = (9x-4)\log_4(3)$$ so that

$$(9x-4)\log_4(4) = (9x-4)\log_4(3).$$

Now can you solve it?

• On the left hand side, do you use the same property? Sorry, I'm having a serious brain fart here. Could you please expand the last step a little more? – Jeel Shah Oct 10 '13 at 12:37
• @gekkostate Does that help clarify? – Neal Oct 10 '13 at 13:24

Note that your equation is equivalent to $$\frac{4^{9x - 4}}{3^{9x-4}} = 1$$ or $$\left(\frac{4}{3}\right)^{9x-4} = 1.$$

• Interesting! I hadn't considered that. – Jeel Shah Oct 10 '13 at 16:00
• (+1) this hint is much more cleaner than the accepted answer and easier to solve IMO – zerosofthezeta Oct 10 '13 at 18:31
• @zerosofthezeta While this is more cleaner, I personally found the other answer more useful as it related the $\log$ properties which is what we are covering currently. – Jeel Shah Oct 11 '13 at 1:09

You want to solve
$$4^Q=3^Q$$ Which becomes

$$\left({4\over3}\right)^Q=1$$ Hence $$Q=0$$

• This is by far the most incisive way to start the problem. – Matt Montag Oct 11 '13 at 7:15
• This is the clearest and most concise description of the correct way to solve the problem. The log function only inserts unnecessary confusion into the mix. The point of the question is to reinforce the idea that anything raised to the power of 0 is 1. – Mark Bailey Oct 11 '13 at 14:33
• @Mark This is not "the" correct way to solve the problem, it is the simplest and quickest way to solve the problem. Rather than inserting "unnecessary confusion into the mix," using logarithms would be pedagogically instructive, as learning how to manipulate logarithms is a necessary component in any elementary algebra class (and the OP intimates in a comment here that the logarithm answer was accepted because it relates best to the material being covered in class - thus your perception of "the point of the question" is plausible but not certain to my mind). – anon Oct 11 '13 at 15:32
• @Mark Either the class is teaching logarithms or it isn't; it is not plausible that OP would accidentally believe class is teaching logarithms when it actually isn't covering them. Manipulating logarithms in this way is the universal textbook way of answering these sorts of questions. While this specific problem admits a shortcut, and even might be designed so that students use the shortcut, manipulating logarithms generally is still a broader lesson and more universal method to be learned, so saying it "inserts unnecessary confusion" and suggesting it's incorrect is off-base in this sense. – anon Oct 11 '13 at 16:29
• Also, the fact that raising numbers to the power of zero yields one cannot alone exhaust the point of the question: we wouldn't solve $2^{3x-1}=3^{2x+1}$ by setting $3x-1=0$ and $2x+1=0$ for example. The point is also that exponentials grow/decay at different rates for different bases, and only intersect thanks to the $a^0=1$ fact. The method of solving by logarithms also illustrates this very same phenomena: the exponents $(\log u)(ax+b)$ and $(\log v)(ax+b)$ increase/decrease at different rates if the scaling factors $u,v$ are distinct, and can only intersect if $ax+b=0$. – anon Oct 11 '13 at 16:37

Note you can take logs to any base for this (provided it is the same both sides).

You get $$(9x-4)\log 4 = (9x-4)\log 3$$

which you can rewrite as $$(9x-4)(\log 4-\log 3)=0$$

In general (for different exponents) you can change the base to $e$ to end up with a simpler equation: $$4^{9x-4}=3^{9x-4}$$ $$\implies e^{(9x-4)\ln4}=e^{(9x-4)\ln3}$$ $$\implies (9x-4)\ln4=(9x-4)\ln3$$ $$\implies x=\frac 49$$

• Although I prefer Hasan's answer to this question, I too think it helpful to convert all exponents and logarithms to base $e$ when unsure how to proceed. – dfeuer Oct 11 '13 at 3:14

Notice, first, that the power of both the numbers is the same. This simplifies matters to a huge extent. You need not take any logarithms.

Your equality is: $$4^{9x-4}=3^{9x-4}$$

You are allowed to multiply/divide numbers with similar powers and take the power of the result. This is one of the Rules of Exponents:

$$\left({x^{\alpha}\over y^{\alpha}}\right)=\left({x\over y}\right)^{\alpha}$$

Doing so, you get:

$$\left({4\over3}\right)^{9x-4}=1$$

Any number raised to power $0$ equals $1$. This means that $9x-4$ should be equal to $0$, or

$$9x-4=0$$

thus, $$x=\frac49$$

This is the solution to the given equality.

$$\begin{array}{rrl} &(9x-4) \log 4 &= (9x-4) \log 3\\ \implies &(9x-4) (\log 4 -\log 3) &= 0\\ \implies &9x -4 &= 0 \mbox{ (as \log 4 \neq \log 3)}\\ \implies &x &= \frac{4}{9} \end{array}$$

For any $A^n = B^n$ where $A \ne B$, $n = 0$

So:

$$9x - 4 = 0$$

$$x = 4/9$$

• I apologize for nitpicking, but “For any $A^n=B^n$” is not a good way to begin a mathematical statement. Also, the meaning of the comma between $A\ne B$ and $n=0$ is not clear. +1 for simplicity of the argument though. – Carsten S Oct 11 '13 at 9:18

So you thought @fasttouch was complex?

$$x = \frac{4\log\frac{4}{3}- 2 \pi ni}{9 \log \frac{4}{3}}, n \in \Bbb Z$$

Gerry said correctly. Can you solve it for $Q$? Now if $Q \in \mathbb{N}^0$ then the only solution is for $Q=0$. Since $4>3$ it follows that for $Q \in \mathbb{N}^0$ and $Q>0$ it is true that $4^Q>3^Q$. Actually we can generalize that and we can say that for $Q>0$ and $Q\in\mathbb{R}$ it is still true.

Now for $Q<0$ you can write your equation in this way $$\frac{1}{4^{|Q|}} = \frac{1}{3^{|Q|}}$$ and being $1/4<1/3$ you get the opposite of what we discussed before. That means that the only solution is $Q=0$ and that means $9x-4=0$ or $x=4/9$ as user1772257 has pointed out.

\begin{align*} 4^{9x-4} &= 3^{9x-4}\\ \log_4(4^{9x-4}) &= \frac{\log_3(3^{9x-4})}{\log_34}\\ 9x-4 &= \frac{9x-4}{\log_34}\\ 9x-4&=0\\ x &= \frac{4}{9} \end{align*}