Logarithmic Question (1 Viewer)

[Katsumi]

Hi All,

I'm currently stuck on a logarithmic question that i've been attempting for most of an hour with 0 success. There are no answers releases yet & any help would be appreciated

Question below

My main point of confusion comes to what rule to apply first. Do i first have to solve Ln(10) & then isolate the logarithmic function by dividing the answer by 3?

Note that i need to find the value "x"

 

[InteGrand]

Hi All,

I'm currently stuck on a logarithmic question that i've been attempting for most of an hour with 0 success. There are no answers releases yet & any help would be appreciated

Question below

My main point of confusion comes to what rule to apply first. Do i first have to solve Ln(10) & then isolate the logarithmic function by dividing the answer by 3?

 

[mreditor16]

3*(logx(7))=ln(10)
logx(7^3)=ln(10)
logx(343)=ln(10)
x^(logx(343))=x^(ln(10))
343=x^(ln(10))
343^(1/ln(10))=(x^(ln(10))^(1/ln(10))
343^(1/ln(10))=x^((ln(10))*(1/ln(10))
343^(1/ln(10))=x^(1)
x=343^(1/ln(10))

 

[D94]

If you know that x^log_xa = a, then after dividing both sides by 3, x^log_x7 = x^ln10/3

So, 7 = x^ln10/3

Raising both sides by 3/ln10 gives x = 7^3/ln10

Do i first have to solve Ln(10)

This is your first misunderstanding. You can't 'solve' ln10 - this is just a number, like the number 200, you can't solve just that number.

You will need to learn the log rules in order to solve any log problems. They are fundamental as they set the laws that govern what you can and can't do.

 

[Katsumi]

So this is the correct logic?

equation provided in question
3logx7 = ln10
a * logbc = logb(c^a)
logx7^3 = ln10
logbc=a is equal to b^a = c
x^ln10 = 7^3
isolate x by raising both sides to 1/x^ln10
x = (7^3)^1/ln10

 

[leehuan]

So this is the correct logic?

equation provided in question
3logx7 = ln10
a * logbc = logb(c^a)
logx7^3 = ln10
logbc=a is equal to b^a = c
x^ln10 = 7^3
isolate x by raising both sides to 1/x^ln10
x = (7^3)^1/ln10

Yep