Logarithmic Help!

[Todd19]

alright I know how to solve an equation with an exponent on one side but what about a function like this?

5^x=4^(x+1)

I know you need to do this.

log5^x=log^(x+1)

Then

xLog5=(x+1)Log4

But then what?
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1974 Standard Beetle - Pacha

http://veedubclassics.blogspot.com/

https://sites.google.com/site/veedubclassics/home

[zozo]

log 5^x = log 4^(x + 1) xlog5 = (x + 1)log4 x log5/log4 = x + 1 x(log5/log4 - 1) = 1 x = 1/(log5/log4 - 1)


So says google.

Just an FYI, you can enter any equation into the google search window and it'll spit out answers.
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[Todd19]

Thank you that was what I was looking for
_________________
1974 Standard Beetle - Pacha

http://veedubclassics.blogspot.com/

https://sites.google.com/site/veedubclassics/home

[gostros]

wolframalpha.com is better IMO.

x=-(2 log(2))/(2 log(2)-log(5))