Question

log5(base)^x + log5(base)^6x+5 =1 Please help

Answer 1

Is this your question?
\[\log_{5} x+ \log_{5} 6x+5=1\]

Answer 2

\[\sqrt{x} \sqrt{6x+5}\] should be added

Answer 3

Is this your question
\[\log_{5} {\sqrt x}+\log_{5} {\sqrt{6x+5}}=1\]
?

Answer 4

yes.

Answer 5

We know that
\[ \log_c a+ \log_c b= \log_c {ab}\]
here we have
\[ \log_5 {\sqrt x}+\log_5 \sqrt{6x+5}= \log_{5} \sqrt {x\times (6x+5)}\]
so we get
\[\log_{5} \sqrt {x\times (6x+5)}=1\]
also
\[\log_{c} a^b=b* \log_{c} {a}\]
so
\[\frac{1}{2}\log_{5} {x\times (6x+5)}=1\]
we get now
\[\log_{5} {x\times (6x+5)}=2\]
\[x*(6x+5)=25
\]
\[6x^2+5x-25=0\]
Can you solve now?

Answer 6

Wow. Yes I can solve from here. I guess I just need to memorized the formulas. We went over this in class last night for the first time. I appreciate your help, thanks.

Answer 7

Welcome:)
Tia1977 you'll get two values for x, your answer is the one which is positive

Answer 8

ok