The Purplemath Forums

by **maggiemagnet** on Mon May 18, 2009 5:26 pm

Solve log₂(log₈(x)) = log₈(log₂(x)).

Because 2³ = 8, I'm wondering if changing bases might help? thank you for your advice.

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by **stapel_eliz** on Fri May 22, 2009 5:48 pm

maggiemagnet wrote:Solve log₂(log₈(x)) = log₈(log₂(x)).

Because 2³ = 8, I'm wondering if changing bases might help? thank you for your advice.

I think you're on the right track. Use the change-of-base formula to convert everything to base-2:

$\log_8(x)\, =\, \frac{\log_2(x)}{\log_2(2^3)}\, =\, \frac{\log_2(x)}{3}$

...so:

$\log_2(\log_8(x))\, =\, \log_2\left(\frac{\log_2(x)}{3}\right)$

Also:

$\log_8(\log_2(x))\, =\, \frac{\log_2(\log_2(x))}{\log_2(2^3)}\, =\, \frac{\log_2(\log_2(x))}{3}$

Using another log rule, you get:

$\frac{\log_2(\log_2(x))}{3}\,=\, \log_2(\sqrt[3]{\log_2(x)})$

$\log_2\left(\frac{\log_2(x)}{3}\right)\, =\,\log_2(\log_2(\sqrt[3]{x}))$

Then:

$\sqrt[3]{\log_2(x)}\, =\, \log_2(\sqrt[3]{x})$

$\sqrt[3]{\log_2(x)}\, =\, \frac{1}{3}\log_2(x)$

$\log_2(x)\, =\, \frac{1}{27}\log_2^3(x)$

$0\, =\, \frac{1}{27}\log_2^3(x)\, -\, \log_2(x)$

$0\, =\, \log_2(x)\left(\frac{1}{27}\log_2^2(x)\, -\, 1\right)$

Then log₂(x) = 0, so x = 1, or (log₂(x))² = 27, log₂(x) = sqrt[27], so x = 2 ^sqrt[27].

Comparing with the original expressions, you can't have x = 1, because then log₂(x) = 0, but you can't take log₈(log₂(1)) = log₈(0).

Whew! I'll leave the check of the other solution value to you. :shock:

by **maggiemagnet** on Sat May 23, 2009 12:45 pm

Goodness! Now I don't feel so bad about bogging down!

Thank you so much! :clap:

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solving log_2(log_8(x)) = log_8(log_2(x))

solving log_2(log_8(x)) = log_8(log_2(x))

Re: solving log_2(log_8(x)) = log_8(log_2(x))