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# logarithm equation

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

Guest Aug 18, 2015

#2
+17711
+5

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

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

Since 10log10 (Anything)  =  Anything

--->   x2 - 2x  =  2x + 12

--->   x2 - 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 log10 from both sides? -- So long as all of both sides are in log form, yes!

geno3141  Aug 18, 2015
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#1
0

Guest Aug 18, 2015
#2
+17711
+5

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

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

Since 10log10 (Anything)  =  Anything

--->   x2 - 2x  =  2x + 12

--->   x2 - 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 log10 from both sides? -- So long as all of both sides are in log form, yes!

geno3141  Aug 18, 2015

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