The discussion revolves around solving the equation 2^(t-1) * 6^(t-2) = 20,000, which involves logarithmic and exponential concepts.
The discussion is ongoing, with participants exploring different approaches to the problem. Some have provided insights into properties of exponents, while others are questioning how to apply these properties in the context of the given equation.
There appears to be confusion regarding the application of exponent properties, particularly when dealing with different coefficients in the terms of the equation.
One possibility is to take the ##\log_2## of both sides of the equation.kolleamm said:Homework Statement
2^(t-1) * 6^(t-2) = 20,000
Homework Equations
The Attempt at a Solution
I have no idea how to solve this, although I do understand the basics of logarithms
Could you find x given axbx=c?kolleamm said:Homework Statement
2^(t-1) * 6^(t-2) = 20,000
Homework Equations
The Attempt at a Solution
I have no idea how to solve this, although I do understand the basics of logarithms
would it beharuspex said:Could you find x given axbx=c?
haruspex said:Could you find x given axbx=c?
This is not even remotely close. You need to review the properties of exponents. A web search on "properties of exponents" or "laws of exponents" should be enlightening.kolleamm said:would it be
log(a^b^c) = x * x ?
One of the properties I found is :Mark44 said:This is not even remotely close. You need to review the properties of exponents. A web search on "properties of exponents" or "laws of exponents" should be enlightening.
That is not the property you need here.kolleamm said:One of the properties I found is :
x^a * x^b = x^(a+b)
but how would I use that property since I have two different coefficient values?
You can write 2t-1=2t * 2-1 and 6t-2=6t * 6-2.kolleamm said:One of the properties I found is :
x^a * x^b = x^(a+b)
but how would I use that property since I have two different coefficient values?