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# Solve X in logarithm equation

+2
33
4
+36

Hello!

I encountered a equation where im having trouble finding x

The equation is \(lg(1,5x+14) + lg2 = 0,5 * lg{x}^{4}\)

First i used the first logarithm law which gave me \(lg(3x + 24) = 0,5 * lg{x}^{4}\)

Then i used the third law which gave me \(lg(3x + 24) = 0,5 * 4 *lgx\)

Then what do i do :P

Thankfull for any help <3

kilander  Sep 14, 2017
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#1
+76102
+2

log (1.5x + 14) + log (2)  =  0.5 log (x^4)

Other than a slight mistake, you were on the right track..... let's go from your last step

log (3x + 28)   =  0.5 * 4 log (x)

log (3x + 28 )  = 2 * log (x)

Let's write the "2"  back as an exponent

log (3x + 28)  =  log (x^2)

Since the we have a single log on each side....we can forget those and solve this :

3x + 28   = x^2    rearranging, we have that

x^2 - 3x  - 28  =0      this is factorable as

(x -7) ( x+4)  = 0

Setting both factors to 0 and solving for x, we have that x  = 7  or  x  = -4

Note that both solutions will solve the simplified form of the problem ,i.e.,  log (3x + 28)  =  log (x^2)    !!!!

CPhill  Sep 14, 2017
#2
+36
+1

Thanks alot for your help :) It just saved me inbefore my exam tomorrow <33

kilander  Sep 14, 2017
#3
+76102
0

No prob.....!!!!

CPhill  Sep 14, 2017
#4
+3
+2

Thanks alot this is so much help.

bdominick  Sep 14, 2017

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