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May 12th, 2019, 05:56 AM   #1
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Change of Base - please verify

Two part question: Use Change of Base to show how to calculate log base 4 (62). Then find its equivalent in log base 2. Round answers to nearest thousandth.

I attempted both parts. Please check my math strategies. Thank you.

PART ONE
(log 62)/(log4)
=(1.79239)/(0.6020599)
=2.977

PART TWO: find equivalent in log base 2

log2 (N) = 2.977

2^2.977 = N
N=7.873

Is this a correct method to solve? Thank you.
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May 12th, 2019, 03:51 PM   #2
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$x=\log_4(62) \implies 4^x=62 \implies (2^2)^x = 62 \implies 2^{2x} = 62 \implies 2x =\log_2(62) \implies x = \dfrac{1}{2} \log_2(62) = \log(62)^{1/2} = \log_2 \sqrt{62}$

$\sqrt{62} \approx 7.874$
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May 12th, 2019, 05:12 PM   #3
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Thank you. I'm going to post little more detail from your steps to quickly clarify for myself if I reference this post again in future:


4^x=62
(2^2)^x=62
(2)^2x=62

b^P=N
log base b (N) = Power

so the (2)^2x=62 becomes
log base 2 (62)=2x

divide both sides by 2

[log base 2 (62)]/2 = 2x/2

0.5 log base 2 (62) = x

use logarithm property power rule: log base b (M^p) = p log base b (M)
rewrite as: log base 2 (62^0.5) = x
62^0.5 is sqrt 62

log base 2 (sqrt 62)

----------------------------
Another method using Change of Base formula to log base 2:
log base 4 (62) = 2.977098155

log (x) / log 2 = 2.977098155
log (x) / 0.3010299957 = 2.977098155 Then cross multiply the decimal and 2.977098155

log base 10 (x) = 0.89616958447

10^0.89616958447 =7.874
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May 12th, 2019, 05:46 PM   #4
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Quote:
Originally Posted by Seventy7 View Post
find its equivalent in log base 2.
That wording is unclear.

I think it's intended to mean "find log$_{_{\large2}}\!$(62)", which is 5.954 to 3 decimal places.
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