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coolaidd
solve. then round answer to the nearest hundredth. 6^x-3=52
\[6^{x-3} = 52\]You will need to use logarithms to solve this one
Specifically, you will need to use the property of logarithms such that will allow you to work with the exponent: So, if you have \[x^y\], if you take the log of this you obtain: \[log_a(x^y) = y * log_a(x)\]The exponent now becomes a factor in the product
For your equation, you will want to take the logarithm of both sides, then you can solve for x. You can use any base for the logarithm: \[log_a\] I'm sure you're familiar with the concept of logarithms, but here is nice way to remember what it does: you can say that when you have the expression: \[log_ab = x\]You are looking for the number you raise 'a' to in order to obtain 'b' is 'x'. \[a^x = b\] \[log_28=3\]The number you raise 2 to in order to obtain the value 8, is 3. \[2^3 = 8\]
Does this seem familiar or helpful?
OK, cool, I see that you had a similar question - http://openstudy.com/users/coolaidd#/updates/4fff44afe4b09082c0707aad
Do you think you can solve this problem?
log_5 x=3 x = 5^3 = 125
6^(x - 3) = 52 (x - 3) ln 6 = ln 52 x - 3 = ln 52/ln 6 x = 3 + ln 52/ln 6 ≈ 5.20523
Sure, good luck with the rest of your studies! Maybe you can help with some questions even ;)