$${log}_{10}\left({{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}\right) = {log}_{10}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}}\right)$$

Guest Aug 18, 2015

#2**+5 **

Since both logs are to the base 10, take both sides of the equation to the power of 10:

10^{log10 (x^2 - 2x} ) = 10^{log10 (2x + 12)}

Since 10^{log10} (Anything) = Anything

---> x^{2} - 2x = 2x + 12

---> x^{2} - 4x - 12 = 0

---> (x - 6)(x + 2) = 0

---> x = 6 or x = -2

In problems like these, you need to check all the answers; they do work!

Could you just cross out the log_{10} from both sides? -- So long as all of both sides are in log form, yes!

geno3141
Aug 18, 2015

#2**+5 **

Best Answer

Since both logs are to the base 10, take both sides of the equation to the power of 10:

10^{log10 (x^2 - 2x} ) = 10^{log10 (2x + 12)}

Since 10^{log10} (Anything) = Anything

---> x^{2} - 2x = 2x + 12

---> x^{2} - 4x - 12 = 0

---> (x - 6)(x + 2) = 0

---> x = 6 or x = -2

In problems like these, you need to check all the answers; they do work!

Could you just cross out the log_{10} from both sides? -- So long as all of both sides are in log form, yes!

geno3141
Aug 18, 2015