Question

What is 2^5x+3=1/8 in exponential form for logarithmic equations?

Answer 1

I'm not sure what you mean, but I think you want the equation in logarithmic form? Because it's already in exponential form.

Answer 2

Sorry.. I mean to say solve this equation using algebraic procedure. Its a logarithmic equation. So here is the equation once again: 2^5x+3=1/8

Answer 3

If that's the case, then do the following:

1) subtract 3 to the other side
2^(5x)=(1/8)-3

2) Take the log base 2 of both sides following the rule y = log2(x), 2^y=x
\[\log_{2} (2^{5x})=\log_{2}(1/8 - 3)\]

3) The log base 2 cancels the 2 in 2^5x leaving you with 5x
\[5x=\log_{2}(1/8-3)\]

4) divide both sides by 5
\[x=\log_2(1/8-3)/5\]

This is a logarithmic equation because it has logarithms in it. This is different from an exponential equation which you provided. In fact, it is the inverse.

Answer 4

But the answer in my textbook is -6/5.. LOL!

Answer 5

I think I wrote the problem wrong.. its 2^(5x+3)=1/8

Answer 6

Okay, so the steps are the same.

1) Take the log base 2 of both sides because your base in the exponential function 2^(5x+3) is 2.
\[\log_{2}(2^{5x+3})=\log_2(1/8) \]

2) Using the logarithm property
\[\log_a(a^{x}) = x\] where x is some function, employ the property on the left hand side of the equation to get\[5x+3=\log_{2}(1/8) \]

3) Another logarithm property states that\[\log_{a}(x/y)=\log_a(x)-\log_a(y) \]where x and y are some function, employ this on the right hand side to get\[5x+3=\log_2(1)-\log_2(8)\]
4) loga(x) = N is equivalent to a^N=x so 2^N=1 when N=0 and 2^N=8 when N=3 giving you\[5x+3=0-3=-3\]
5) Subtract 3 on both sides of the equation and divide both sides of the equation by 5 to get\[x=(-3-3)/5 = -6/5\]

Answer 7

Thank you so much for your help. :)