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Solve for x :S

Solve for x:



Where do I even start? :s Bases are different.

I tried exponentiating both sides (raising both sides to the power of x and then also tried the same but to the power of 8, also tried expressing the whole expression in terms of e, rather than 8. Got somewhere but nowhere useful, I believe.

Thanks!

Sid
Original post by Sidhant Shivram
...


Can't see the image.

Guessing - you're going to need to change the base.
Original post by Sidhant Shivram
Solve for x:



Where do I even start? :s Bases are different.

I tried exponentiating both sides (raising both sides to the power of x and then also tried the same but to the power of 8, also tried expressing the whole expression in terms of e, rather than 8. Got somewhere but nowhere useful, I believe.

Thanks!

Sid

As ghostwalker said, the picture isn't there. However, it may be useful to remember that log⁡(ab)=blog⁡(a)\log(a^b) = b \log(a).
Original post by ghostwalker
Can't see the image.

Guessing - you're going to need to change the base.


Original post by Smaug123
As ghostwalker said, the picture isn't there. However, it may be useful to remember that log⁡(ab)=blog⁡(a)\log(a^b) = b \log(a).




18.b) The 2nd part
EX2Q18b.png7.4KB
EXQ18.b BIG.png9.2KB
Original post by Sidhant Shivram


18.b) The 2nd part

log⁡x(8)=log⁡e(8)log⁡e(x)=log⁡e(81/log⁡e(8))\log_x(8) = \frac{\log_e(8)}{\log_e(x)} = \log_e(8^{1/\log_e(8)}) unless x≤0x \leq 0 or x=1x=1. That's how I'd end up doing it, I think.
Original post by Smaug123
log⁡x(8)=log⁡e(8)log⁡e(x)=log⁡e(81/log⁡e(8))\log_x(8) = \frac{\log_e(8)}{\log_e(x)} = \log_e(8^{1/\log_e(8)}) unless x≤0x \leq 0 or x=1x=1. That's how I'd end up doing it, I think.


ImageUploadedByTapatalk1419699803.988034.jpg

Ah yes. Makes sense. I used your method. Got some sensible answers. However, the answers given are : 4 and 16 x root2 :/


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you can write the 2 on its own as 2log33
Original post by the bear
you can write the 2 on its own as 2log33

Exactly the same as what I was thinking.
May be better to write it as the logarithm of 9 to base 3 however.
Original post by the bear
you can write the 2 on its own as 2log33


Original post by MathMeister
Exactly the same as what I was thinking.
May be better to write it as the logarithm of 9 to base 3 however.


The first part is fine. its the 2nd part, part (b), i.e., thats bothering me.
Original post by Sidhant Shivram
The first part is fine. its the 2nd part, part (b), i.e., thats bothering me.

Change of base rule.
Original post by Smaug123
log⁡x(8)=log⁡8(8)log⁡8(x)\log_x(8) = \frac{\log_8(8)}{\log_8(x)} .


this is easier
Original post by MathMeister
Change of base rule.


Original post by the bear
this is easier


attachment.php ?

Got some sensible answers. However, the answers given are : 4 and 16 x root2 :/
Original post by Sidhant Shivram
attachment.php ?

Got some sensible answers. However, the answers given are : 4 and 16 x root2 :/

Haven't got a pen or pencil (or paper) on me atm but by just looking at the Q-maybe use the change of base rule and try to form a quadratic? (remembering the logarithmic function is defined for positive values of x.
I may be wrong however as I have not tried it out myself.
Original post by Sidhant Shivram
...


The answers given check out.

Your fourth line of working does not follow from third.
Original post by lazy_fish
The answers given check out.

Your fourth line of working does not follow from third.


Oops! Got it! thank you so much!

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