I need to solve the equation : $\ln(x+2)+\ln(5)=\lg(2x+8)$

With the change of base formula we can turn this into: $\ln(x+2)+\ln(5)=\frac{\ln(2x+8)}{\ln(10)}$

We can also simplify the LHS with the product rule so: $\ln(5(x+2))=\frac{\ln(2x+8)}{\ln(10)}$

Solving the fraction gives us: $\ln(10) \, \ln(5(x+2)) = \ln(2x+8)$

Simplifying the LHS even further: $\ln(5x+10)^{\ln(10)}=\ln(2x+8)$

We can then see that $(5x+10)^{\ln(10)}=2x+8$

And this is where I get stuck, I can't seem to figure out how to expand this term. Does anyone know how to solve this?

  • $\begingroup$ What makes you think that it can be solved analytically? $\endgroup$ – Kaster Aug 9 '15 at 17:51
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    $\begingroup$ It's quite possible there's a mistake in the problem. I don't think you'll be able to solve that for an exact answer. $\endgroup$ – Mike Aug 9 '15 at 17:52
  • $\begingroup$ What is $lg(x)$? $\endgroup$ – Billy Rubina Aug 9 '15 at 20:25
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    $\begingroup$ @voyska I've seen $\lg x$ represent a logarithm of base $2$. However, in this case, it appears to be an error edited in by someone else. The original problem before edits had $\log(2x+8)$. And the OP's work treated it as a common logarithm. $\endgroup$ – Mike Aug 9 '15 at 20:55

You can find answer by researching of graphic of functions,I don't think you can get it by algebric methodenter image description here


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