5

So this is what I have done so fare

$$2 \log _3(x)+\log _9(x)=10$$

I know that

$$\log _9(x)=\log _3\left(\sqrt{x}\right)$$

I therefore have

$$\log _3\left(x^{5/2}\right)=10$$

However here is where I realise that I have not properly understood the underlying of log. Since I know that the answer is 81 I was able to realise that

$$\log _3(x)=4$$

Is the same as x==3^4, but I do not understand if Im dividing or multiplying or what is going on. I thought that

$$\log _3(x)$$

meant that

$$3^?=x$$

Basically trying to find the exponent? At least that is the dummed version that I got thought in class

Could someone explain what is going on and what I do not understand , in such a way that I am able to grasp the concept.

Thomas Andrews
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ALEXANDER
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    One way to look at this is that $\log_3(x) = 2 \log_9(x)$, since if $3^y = x$, then $9^{\frac{y}{2}} = 3^y = x$. This then gives you that $4\log_9(x) + \log_9(x) = 10$, or $\log_9(x) = 2$. You can then exponentiate both sides, giving $9^{\log_9(x)} = 9^2$, or $x = 9^2 = 81$. – qaphla Aug 20 '14 at 21:55
  • If you have already seen the laws of logarithms, you can rewrite $\log_{3}(x^{\frac{5}{2}})$ as $\frac{5}{2}\log_{3}x$. – user84413 Aug 20 '14 at 22:03

4 Answers4

4

You correctly got:$$\log_3(x^{5/2})=10$$next step gives you:$$x^{5/2}=3^{10}$$therefore:$$x=(3^{10})^{2/5}$$Hopefully you can complete from here

Mufasa
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$$\log _3\left(x^{5/2}\right)=10$$

I will use the approach that you stated to find the answer.

You are on the right track when you state that $\log_3(x)$ is defined to be the solution to the equation $3^?=x$ where we solve for question-mark. In our case, the question mark is already known and we have to solve for $x$. For the equation above, question-mark equals 10. That's what the equals sign tells you in this scenario.

Now, we have to solve the following equation for $x$.

$$3^{10} = x^{5/2}$$

user157227
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2

Perhaps if you cleared out the denominator first it would be clearer what is going on:

$2\log_{3}x+\frac{1}{2}\log_3{x}=10\\ 4\log_{3}x+\log_3{x}=20\\ 5\log_{3}x=20\\ \log_3 x = 4\\ x=3^4$

Does this help?

Paul Sundheim
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1

Basically, you have a function $f:\mathbb{R}\to\mathbb{R}^+$ defined by: $$ f(x)=3^x.$$ Since $f$ is increasing and surjective, there exists a function $g:\mathbb{R^+}\to \mathbb{R}$ that is the inverse of $f$, i.e. satisfies: $$\forall x\in\mathbb{R},\quad g(f(x))=x,$$ $$\forall x\in\mathbb{R}^+,\quad f(g(x))=x.$$ We call such function $\log_3$, i.e. $\log_3(x)=g(x)$. So yes, if $x\in\mathbb{R}^+$ and $$ 3^{?}=x, $$ we have: $$ ? = \log_3(x).$$ All the algebraic properties of a logarithm function can be derived from the properties of its inverse function. For instance, $f(x)f(y)=f(x+y)$ gives $g(xy)=g(x)+g(y)$. This also gives that if you pick a "reference logarithm" as the inverse function of $e^x$ and you call it simply $\log x$, then: $$\log_a(x) = \frac{\log x}{\log a},$$ so: $$\log_{a^2}(x) = \frac{\log x}{\log a^2} = \frac{\log x}{2\log a} = \frac{1}{2}\log_a(x).$$ This properties give that your equation is equivalent to: $$\left(2+\frac{1}{2}\right)\log_3(x) = 10, $$ or to: $$\log_3(x) = 4,$$ so $x=3^4$ as you correctly stated.

Jack D'Aurizio
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