# I am trying to proof the following expression (without a calculator of course). log_(12)18xxlog_(24)54+5(log_(12)18−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 log_(12)3 and 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.

Proof the expession ${\mathrm{log}}_{12}18×{\mathrm{log}}_{24}54+5\left({\mathrm{log}}_{12}18-{\mathrm{log}}_{24}54\right)=1$
I am trying to proof the following expression (without a calculator of course).
${\mathrm{log}}_{12}18×{\mathrm{log}}_{24}54+5\left({\mathrm{log}}_{12}18-{\mathrm{log}}_{24}54\right)=1$
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)
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honotMornne
Let ${\mathrm{log}}_{12}18=a,\phantom{\rule{thinmathspace}{0ex}}{\mathrm{log}}_{24}54=b$. So you want to prove that $ab+5\left(a-b\right)=1$
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:
$2a-1=\left(2-a\right){\mathrm{log}}_{2}3\phantom{\rule{2em}{0ex}}3b-1=\left(3-b\right){\mathrm{log}}_{2}3.$
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.
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Sonia Elliott
The follow-your-nose approach would be, in my opinion,
$\begin{array}{rl}{\mathrm{log}}_{12}18\cdot {\mathrm{log}}_{24}54& +5\left({\mathrm{log}}_{12}18-{\mathrm{log}}_{24}54\right)\\ & =\frac{\mathrm{log}18}{\mathrm{log}12}\frac{\mathrm{log}54}{\mathrm{log}24}+5\left(\frac{\mathrm{log}18}{\mathrm{log}12}-\frac{\mathrm{log}54}{\mathrm{log}24}\right)\\ & =\frac{\mathrm{log}18}{\mathrm{log}12}\frac{\mathrm{log}54}{\mathrm{log}24}+5\left(\frac{\mathrm{log}18\cdot \mathrm{log}24-\mathrm{log}54\cdot \mathrm{log}12}{\mathrm{log}12\cdot \mathrm{log}24}\right)\\ & =\frac{\mathrm{log}18\cdot \mathrm{log}54+5\mathrm{log}18\cdot \mathrm{log}24-5\mathrm{log}54\cdot \mathrm{log}12}{\mathrm{log}12\cdot \mathrm{log}24}.\end{array}$
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
$\begin{array}{rl}& \frac{\mathrm{log}18\cdot \mathrm{log}54+5\mathrm{log}18\cdot \mathrm{log}24-5\mathrm{log}54\cdot \mathrm{log}12}{\mathrm{log}12\cdot \mathrm{log}24}\\ & \phantom{\rule{2em}{0ex}}=\frac{\left(t+2h\right)\left(t+3h\right)+5\left(t+2h\right)\left(3t+h\right)-5\left(t+3h\right)\left(2t+h\right)}{\left(2t+h\right)\left(3t+h\right)}\\ & \phantom{\rule{2em}{0ex}}=\frac{{t}^{2}+5th+6{h}^{2}+15{t}^{2}+35th+10{h}^{2}-\left(10{t}^{2}+35th+15{h}^{2}\right)}{6{t}^{2}+5th+{h}^{2}}=1.\end{array}$