Cyrus Travis

2022-10-22

Proof the expession ${\mathrm{log}}_{12}18\times {\mathrm{log}}_{24}54+5({\mathrm{log}}_{12}18-{\mathrm{log}}_{24}54)=1$

I am trying to proof the following expression (without a calculator of course).

${\mathrm{log}}_{12}18\times {\mathrm{log}}_{24}54+5({\mathrm{log}}_{12}18-{\mathrm{log}}_{24}54)=1$

I know this isn't a difficult task but it's just killing me. I have tried many things, among which was base transformation to 12 and expressing every logarithm in terms of ${\mathrm{log}}_{12}3$ and ${\mathrm{log}}_{12}2$ but every time I try to do it, I mess up something. I don't know if my concentration is terrible or I'm doing something wrong.

Thanks ;) ( if there are more levels to this task, I'd like a hint, not a complete solution)

I am trying to proof the following expression (without a calculator of course).

${\mathrm{log}}_{12}18\times {\mathrm{log}}_{24}54+5({\mathrm{log}}_{12}18-{\mathrm{log}}_{24}54)=1$

I know this isn't a difficult task but it's just killing me. I have tried many things, among which was base transformation to 12 and expressing every logarithm in terms of ${\mathrm{log}}_{12}3$ and ${\mathrm{log}}_{12}2$ but every time I try to do it, I mess up something. I don't know if my concentration is terrible or I'm doing something wrong.

Thanks ;) ( if there are more levels to this task, I'd like a hint, not a complete solution)

honotMornne

Beginner2022-10-23Added 12 answers

Let ${\mathrm{log}}_{12}18=a,\phantom{\rule{thinmathspace}{0ex}}{\mathrm{log}}_{24}54=b$. So you want to prove that ${ab+5(a-b)=1}$

Then you get ${12}^{a}=18$ and ${24}^{b}=54$. Now factor everything in powers of $2$ and $3$ to get

${2}^{2a-1}={3}^{2-a}\phantom{\rule{2em}{0ex}}{2}^{3b-1}={3}^{3-b}.$

From this taking $\mathrm{log}$ base $2$ you will get:

$2a-1=(2-a){\mathrm{log}}_{2}3\phantom{\rule{2em}{0ex}}3b-1=(3-b){\mathrm{log}}_{2}3.$

Furthermore you get

$\frac{2a-1}{2-a}=\frac{3b-1}{3-b}.$

Now simplify this and see you will get the expression written on the first line.

Then you get ${12}^{a}=18$ and ${24}^{b}=54$. Now factor everything in powers of $2$ and $3$ to get

${2}^{2a-1}={3}^{2-a}\phantom{\rule{2em}{0ex}}{2}^{3b-1}={3}^{3-b}.$

From this taking $\mathrm{log}$ base $2$ you will get:

$2a-1=(2-a){\mathrm{log}}_{2}3\phantom{\rule{2em}{0ex}}3b-1=(3-b){\mathrm{log}}_{2}3.$

Furthermore you get

$\frac{2a-1}{2-a}=\frac{3b-1}{3-b}.$

Now simplify this and see you will get the expression written on the first line.

Sonia Elliott

Beginner2022-10-24Added 4 answers

The follow-your-nose approach would be, in my opinion,

$\begin{array}{rl}{\mathrm{log}}_{12}18\cdot {\mathrm{log}}_{24}54& +5({\mathrm{log}}_{12}18-{\mathrm{log}}_{24}54)\\ & =\frac{\mathrm{log}18}{\mathrm{log}12}\frac{\mathrm{log}54}{\mathrm{log}24}+5{\textstyle (}\frac{\mathrm{log}18}{\mathrm{log}12}-\frac{\mathrm{log}54}{\mathrm{log}24}{\textstyle )}\\ & =\frac{\mathrm{log}18}{\mathrm{log}12}\frac{\mathrm{log}54}{\mathrm{log}24}+5{\textstyle (}\frac{\mathrm{log}18\cdot \mathrm{log}24-\mathrm{log}54\cdot \mathrm{log}12}{\mathrm{log}12\cdot \mathrm{log}24}{\textstyle )}\\ & =\frac{\mathrm{log}18\cdot \mathrm{log}54+5\mathrm{log}18\cdot \mathrm{log}24-5\mathrm{log}54\cdot \mathrm{log}12}{\mathrm{log}12\cdot \mathrm{log}24}.\end{array}$

To save space, write $t=\mathrm{log}2$ and $h=\mathrm{log}3$. Then $\mathrm{log}18=\mathrm{log}2+2\mathrm{log}3=t+2h$, etc., and we get

$\begin{array}{rl}& \frac{\mathrm{log}18\cdot \mathrm{log}54+5\mathrm{log}18\cdot \mathrm{log}24-5\mathrm{log}54\cdot \mathrm{log}12}{\mathrm{log}12\cdot \mathrm{log}24}\\ & \phantom{\rule{2em}{0ex}}=\frac{(t+2h)(t+3h)+5(t+2h)(3t+h)-5(t+3h)(2t+h)}{(2t+h)(3t+h)}\\ & \phantom{\rule{2em}{0ex}}=\frac{{t}^{2}+5th+6{h}^{2}+15{t}^{2}+35th+10{h}^{2}-(10{t}^{2}+35th+15{h}^{2})}{6{t}^{2}+5th+{h}^{2}}=1.\end{array}$

$\begin{array}{rl}{\mathrm{log}}_{12}18\cdot {\mathrm{log}}_{24}54& +5({\mathrm{log}}_{12}18-{\mathrm{log}}_{24}54)\\ & =\frac{\mathrm{log}18}{\mathrm{log}12}\frac{\mathrm{log}54}{\mathrm{log}24}+5{\textstyle (}\frac{\mathrm{log}18}{\mathrm{log}12}-\frac{\mathrm{log}54}{\mathrm{log}24}{\textstyle )}\\ & =\frac{\mathrm{log}18}{\mathrm{log}12}\frac{\mathrm{log}54}{\mathrm{log}24}+5{\textstyle (}\frac{\mathrm{log}18\cdot \mathrm{log}24-\mathrm{log}54\cdot \mathrm{log}12}{\mathrm{log}12\cdot \mathrm{log}24}{\textstyle )}\\ & =\frac{\mathrm{log}18\cdot \mathrm{log}54+5\mathrm{log}18\cdot \mathrm{log}24-5\mathrm{log}54\cdot \mathrm{log}12}{\mathrm{log}12\cdot \mathrm{log}24}.\end{array}$

To save space, write $t=\mathrm{log}2$ and $h=\mathrm{log}3$. Then $\mathrm{log}18=\mathrm{log}2+2\mathrm{log}3=t+2h$, etc., and we get

$\begin{array}{rl}& \frac{\mathrm{log}18\cdot \mathrm{log}54+5\mathrm{log}18\cdot \mathrm{log}24-5\mathrm{log}54\cdot \mathrm{log}12}{\mathrm{log}12\cdot \mathrm{log}24}\\ & \phantom{\rule{2em}{0ex}}=\frac{(t+2h)(t+3h)+5(t+2h)(3t+h)-5(t+3h)(2t+h)}{(2t+h)(3t+h)}\\ & \phantom{\rule{2em}{0ex}}=\frac{{t}^{2}+5th+6{h}^{2}+15{t}^{2}+35th+10{h}^{2}-(10{t}^{2}+35th+15{h}^{2})}{6{t}^{2}+5th+{h}^{2}}=1.\end{array}$

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