Logarithms

[123]

Hello, i'm not sure for this
\(\displaystyle log_2^{2}(x+1)+log_\frac{1}{2}(x+1)=6\)
i do like this \(\displaystyle log_2(x+1)=a\), and have \(\displaystyle a^2+\frac{1}{a}-6=0\), and then \(\displaystyle a^3-6a+1=0\).. It's good or not? I can't solve it.

 

[Deleted member 4993]

Hello, i'm not sure for this
\(\displaystyle log_2^{2}(x+1)+log_\frac{1}{2}(x+1)=6\)
i do like this \(\displaystyle log_2(x+1)=a\), and have \(\displaystyle a^2+\frac{1}{a}-6=0\), and then \(\displaystyle a^3-6a+1=0\).. It's good or not? I can't solve it.

almost correct!

\(\displaystyle log_{\frac{1}{2}}(1+x) \ = \ -log_2(1+x) \ = \ -a\)

so your equation is:

\(\displaystyle a^2 \ - \ a \ - \ 6 \ = \ 0\)

 

[123]

Hello, i'm not sure for this
\(\displaystyle log_2^{2}(x+1)+log_\frac{1}{2}(x+1)=6\)
i do like this \(\displaystyle log_2(x+1)=a\), and have \(\displaystyle a^2+\frac{1}{a}-6=0\), and then \(\displaystyle a^3-6a+1=0\).. It's good or not? I can't solve it.

almost correct!

\(\displaystyle log_{\frac{1}{2}}(1+x) \ = \ -log_2(1+x) \ = \ -a\)

so your equation is:

\(\displaystyle a^2 \ - \ a \ - \ 6 \ = \ 0\)

Oh, really. Thank Subhotosh Khan!