Question 312696
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I'm presuming you mean:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\,\cdot\,6^{4x+3}\ -\ 15\ =\ 20]


Add 15 to both sides and divide both sides by 5:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6^{4x+3}\ =\ 7]


Take the log of both sides -- but to which base?  Actually, in the final analysis, it doesn't matter.  The only thing that could possibly influence your decision would be if you have a calculator that can find the log to ANY valid base.  I don't know if the TI-84 will do that or not.  MS Excel will do it if you know how.  Assuming that you only have ln and log (base 10) available, pick one and proceed:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ln\left(6^{4x+3}\right)\ =\ \ln(7)]


Use:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_b(x^n)\ =\ n\log_b(x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (4x\,+\,3)\ln\left(6\right)\ =\ \ln(7)]


Divide by *[tex \Large \ln(6)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (4x\,+\,3)\ =\ \frac{\ln(7)}{\ln(6)}]


Add -3 and divide by 4:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\frac{\ln(7)}{\ln(6)}\ -\ 3}{4}]


Is the exact answer -- use your calculator for a numeric approximation.


Now if you actually have the ability to take the log to any base, you can take the base 6 log, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log_6\left(6^{4x+3}\right)\ =\ \log_6(7)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (4x\,+\,3)\log_6\left(6\right)\ =\ \log_6(7)]


But *[tex \Large \log_6(6)\ =\ 1], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (4x\,+\,3)\ =\ \log_6(7)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{\log_6(7)\ -\ 3}{4}]


A somewhat simpler calculation.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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