Logarithms: 4^x = 5(2^(x-1) - 1, 2^x - 3 + 2(2^-x) = 0,

[courteous]

I'm left with three logarithms unsolved :evil:
1) \(\displaystyle 4^x=5*2^{x-1}-1\)

2) \(\displaystyle 2^x-3+2*2^{-x}=0\)

3) \(\displaystyle 8x=x^{\log_8{x^{12}}}\)

four, actually... :twisted:

4) \(\displaystyle \log_4{x}-1=\frac{2}{\log_4{x}}\)

 

[skeeter]

Re: Logarithms

1. change to ...

\(\displaystyle (2^x)^2 - \frac{5}{2} \cdot 2^x + 1 = 0\)

now use the quadratic formula to solve for 2[sup:3s4xp7w7]x[/sup:3s4xp7w7].

2. multiply every term by 2[sup:3s4xp7w7]x[/sup:3s4xp7w7] ...

\(\displaystyle (2^x)^2 - 3 \cdot 2^x + 2 = 0\)

same quadratic drill as #1.

3. \(\displaystyle x^{\log_8{x}^{12}} = x^{12\log_8{x}} = \left(x^{\log_8{x}}\right)^{12}\)

now, what does \(\displaystyle x^{\log_8{x}} = ?\)

4. multiply every term by \(\displaystyle \log_4{x}\) ...

\(\displaystyle (\log_4{x})^2 - \log_4{x} = 2\)

what kind of equation? (take another look at #1 and #2)

 

[courteous]

Re: Logarithms

now, what does \(\displaystyle x^{\log_8{x}} = ?\)
what does \(\displaystyle x^{\log_8{x}}\) equal to?

 

[skeeter]

my mistake ... had something else on my mind.

try this ...

\(\displaystyle 8x = x^{\log_8{x^{12}}}\)

\(\displaystyle 8x = x^{12\log_8{x}}\)

\(\displaystyle log_8\) both sides ...

\(\displaystyle \log_8{8x} = \log_8\left[x^{12\log_8{x}}\right]\)

\(\displaystyle \log_8{8} + \log_8{x} = 12\log_8{x} \cdot \log_8{x}\)

\(\displaystyle 0 = 12(log_8{x})^2 - \log_8{x} - 1\)

another quadratic.