logarithms

Albi said:
If log(a)x=2, log(b)x=3 and log(c)x=6 then how much is log(abc)x going to be?
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\(\displaystyle Log_{abc}x = \frac{1}{Log_x(abc)} \ \ = \frac{1}{Log_x(a) + Log_x(b) + Log_x(c)}\)

continue....
 
Albi said:
I think i found the answer and it is 11
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No. Please show your work so we can see what you did. (Or maybe you didn't mean to type 11?)

In the work you did show, you didn't write equations at all!

If [MATH]\log_a(x) = 2[/MATH], then [MATH]\frac{1}{\log_x(a)} = 2[/MATH], so [MATH]\log_x(a) = 1/2[/MATH]. Do that for the other three expressions, and proceed.
 
Dr.Peterson said:
No. Please show your work so we can see what you did. (Or maybe you didn't mean to type 11?)

In the work you did show, you didn't write equations at all!

If [MATH]\log_a(x) = 2[/MATH], then [MATH]\frac{1}{\log_x(a)} = 2[/MATH], so [MATH]\log_x(a) = 1/2[/MATH]. Do that for the other three expressions, and proceed.
Click to expand...
20200828_161116.jpg

Is it correct?
 
Well done for solving this! In case you want to see an alternative method (not a better method, just different)...

Cubist said:
Don't we just need to extend @pka 's post#8 a bit by writing...

[math]x = a^2 = b^3 = c^6 = (abc)^y[/math] and then find y (which is the answer)

Start by writing "a" and "b" in terms of c, then you can substitute these values into [math](abc)^y = c^6[/math] to find y
Click to expand...

Note that a, b, and c are real and +ve since they are used as log bases. Using the post above...

[math]a^2=c^6 ⇒ a=c^3[/math]
[math]b^3=c^6 ⇒ b=c^2[/math]
Sub the above values for "a" and "b" into [math]c^6=(abc)^y[/math]
[math]= \left(c^3 \cdot c^2 \cdot c\right)^y[/math]
[math]=\left(c^6\right)^y[/math]
And since [math]c^6=\left(c^6\right)^y[/math] we can conclude that [math]y=1[/math]
 
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