Multiple exponents for logarithms

[kolleamm]

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

 

[Mark44]

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

One possibility is to take the $\log_2$ of both sides of the equation.

 

[haruspex]

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 a^xb^x=c?

 

[kolleamm]

Could you find x given a^xb^x=c?

would it be

log(a^b^c) = x * x ?

 

[Mark44]

Could you find x given a^xb^x=c?

would it be

log(a^b^c) = x * x ?

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]

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.

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?

 

[haruspex]

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?

That is not the property you need here.
Let me ask a different question... how else might you write (xy)^a?

 

[ehild]

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 2^t-1=2^t * 2^-1 and 6^t-2=6^t * 6^-2.