## rosedewittbukater 2 years ago Algebra 2 help and explanation? I get confused when the base of the log isn't 10. Solve the equation and round to the nearest ten-thousandth. log2 4x = 5 (2 is the base)

1. rosedewittbukater

And I need help with 2 log3 x = 54

2. whpalmer4

\[log_2{4x} = 5\] Raise both sides to the 5th power of the log base to get rid of the log: \[4x = 2^5\] \[x= 32/4 = 8\] Check our work: \[log_2{(4*8)} = 5\] \[2^5 = 32 = 4*8\] Remember the logarithm base n of x is just the number to which you raise n to get x. So, the log base 2 of 32 is 5, because 2^5 = 32.

3. whpalmer4

\[2\log_3 x = 54\] I'd divide both sides by 2 to get \[\log_3 x = 27\] Can you tell me what x is now?

4. whpalmer4

Or is the equation \[2\log{(3x)}=54\]?

5. whpalmer4

I'm suspicious that the second problem isn't correct...

6. rosedewittbukater

@whpalmer4 thanks so much! The second one has a base of 3.

7. rosedewittbukater

for the second one I changed it to log3 x^2 = 54 and then 3^54 = x^2 3^27 = x x = 7.625 times 10^12

8. whpalmer4

I would check your rounding carefully on that last one, you've gotten the right number, it would be a shame to still get it wrong (I'm not taking a position either way)

9. musiclover101

it would be|dw:1357961835423:dw|

10. musiclover101

and so the answer is 8 if you solve for x