Logarithms

[IloveManUtd]

Please help me solve this equation:

Solve the simultaneous equations:
xy= 27, logx y + logy x = 2.5

I'm not very good in changing bases, so please help someone. Thx

 

[mmm4444bot]

I'm not very good in changing bases

Then make an educated guess, instead.

Assume that this exercise is cooked up so that x and y are both Whole numbers.

The factors of 27 are 1*27 or 3*9.

log[sub:372186fg]1[/sub:372186fg](27) does not exist, so x and y must be 3 and 9.

Now check. Is the following equation a true statement?

log[sub:372186fg]3[/sub:372186fg](9) + log[sub:372186fg]9[/sub:372186fg](3) = 5/2

If it is, you're done.

 

[mmm4444bot]

While I was eating dinner, I thought of another reasoned-approach that does not involve changing bases.

That sum of logarithms is a sum of exponents, since logarithms represent exponents.

5/2 could be the mixed number 2 + 1/2.

Well, those two numbers as exponents represent squaring (2) and taking the square root (1/2), so it seems reasonable to decompose 5/2 into 2 + 1/2 and think of the following.

log[sub:3iia9jg7]x[/sub:3iia9jg7](y) + log[sub:3iia9jg7]y[/sub:3iia9jg7](x) = 2 + 1/2

So we could assume that:

log[sub:3iia9jg7]x[/sub:3iia9jg7](y) = 2

Switching to exponential form means:

x^2 = y

But y = 27/x, so we would have:

x^2 = 27/x

That easily leads to x = 3, which makes y obvious.

Because switching x and y in the two given equations does not change the original system, we know that assuming log[sub:3iia9jg7]x[/sub:3iia9jg7](y) equals 1/2, instead, will lead to the same solution.

Of course, we should still check our solution. Is the following equation a true statement?

log[sub:3iia9jg7]3[/sub:3iia9jg7](9) + log[sub:3iia9jg7]9[/sub:3iia9jg7](3) = 5/2

If it is, we're done.

 

[soroban]

Hello, IloveManUtd!

I too solved it by "eyeballing" it . . . \(\displaystyle \{x,y\} \,=\,\{3,9\}\)

I solved it algebraically . . . but there must be a better way!

\(\displaystyle \text{Solve: }\;\begin{Bmatrix}xy &=& 27 & [1] \\ \log_xy + \log_y x &=& \frac{5}{2} & [2] \end{Bmatrix}\)

With equation [1], take logs, base \(\displaystyle x\) :

. . \(\displaystyle \log_x(xy) \:=\:\log_x27 \quad\Rightarrow\quad \underbrace{\log_xx}_{\text{This is 1}} +\log_xy \:=\:\log_x27 \quad\Rightarrow\quad \log_xy \:=\:\log_x27-1\quad[3]\)


\(\displaystyle \text{Since }\:\log_yx \:=\:\frac{1}{\log_xy}\)

. . \(\displaystyle \text{ [2] becomes: }\;\log_xy + \frac{1}{\log_xy} \;=\;\frac{5}{2} \quad\Rightarrow\quad 2(\log_xy)^2 - 5(\log_xy) - 2 \:=\:0\)


\(\displaystyle \text{Factor: }\\log_xy - 2)(2\log_xy - 1) \:=\:0\)



\(\displaystyle \text{And we have two equations to solve:}\)


\(\displaystyle \log_xy - 2 \:=\:0 \quad\Rightarrow\quad \log_xy \:=\:2\)

. . \(\displaystyle \text{Substitute [3]: }\;\log_x27-1 \:=\:2 \quad\Rightarrow\quad \log_x27 \:=\:3 \quad\Rightarrow\quad x^3 \:=\:27\)

. . \(\displaystyle \text{Therefore: }\;\boxed{x \,=\,3,\;y\,=\,9}\)


\(\displaystyle 2\log_xy - 1 \:=\:0 \quad\Rightarrow\quad \log_xy \:=\:\tfrac{1}{2}\)

. . \(\displaystyle \text{Substitute [3]: }\;\log_x27-1 \:=\:\tfrac{1}{2} \quad\Rightarrow\quad \log_x27 \:=\:\tfrac{3}{2} \quad\Rightarrow\quad x^{\frac{3}{2}} \:=\:27\)

. . \(\displaystyle \text{Therefore: }\;\boxed{x\,=\,9,\;y\,=\,3}\)

 

[BigGlenntheHeavy]

\(\displaystyle Given: \ xy \ = \ 27 \ and \ log_y(x)+log_x(y) \ = \ \frac{5}{2}, \ x \ > \ 0, \ y \ >0, \ Why?\)

\(\displaystyle Hence, \ log_y(x)+\frac{1}{log_y(x)} \ = \ \frac{5}{2}\)

\(\displaystyle 2[log_y(x)]^2+2 \ = \ 5log_y(x)\)

\(\displaystyle 2[log_y(x)]^2-5log_y(x)+2 \ = \ 0\)

\(\displaystyle [2log_y(x)-1][log_y(x)-2] \ = \ 0\)

\(\displaystyle Ist \ one: \ 2log_y(x) \ = \ 1, \ \implies \ log_y(x) \ = \ 1/2, \ \implies \ x \ = \ y \ ^{1/2}\)

\(\displaystyle Ergo, \ y^{3/2} \ = \ 27, \ \implies \ y \ = \ 9, \ \implies \ x \ = \ 3\)

\(\displaystyle Second \ one: \ x \ = \ 9, \ y \ = \ 3, \ I'll \ leave \ the \ check \ up \ to \ whoever.\)

 

[Deleted member 4993]

Let's assume:

x = y[sup:226lbwwi]m[/sup:226lbwwi] ? y = x[sup:226lbwwi]1/m[/sup:226lbwwi]

log[sub:226lbwwi]x[/sub:226lbwwi]y + log[sub:226lbwwi]y[/sub:226lbwwi]x = 2.5

1/m + m = 2.5

2m[sup:226lbwwi]2[/sup:226lbwwi] - 5m + 2 = 0

(m - 2)(2m - 1) = 0

m = 2 ? y[sup:226lbwwi]3[/sup:226lbwwi] = 27 ? y = 3

and so on.....