Express each of the given question in logarithmic form.
i) e^2x = 15
ii) e^x-4 = 1/2
iii) a^x+1 = 5
Solutions
But am not sure if this is the way it should be done.
i) e^2x = 15
In (15) = 2x
ii) e^x-4 = 1/2
In (1/2) = x-4
iii) a^x+1 = 5
Log 5 with base "a" = x+1
ln e^(2x)
= 2x = ln (15)
next two lack parentheses so I do not know if you mean
e^(x-4) or e^x - 4
If you just left the parens out, the bunch are right
Ok and thanks alot. But those two are without parentheses.
oh, well if e^x - 4 = 1/2
then e^x = 4.5
x = ln 4.5
but I wonder if that is what you mean
your book does not need parentheses as it does not need ^ sign for expoinents.
However on computer keyboard you must be more careful.
To express each of the given equations in logarithmic form, you can use the fact that logarithms are inverses of exponential functions. The logarithmic form for each equation is as follows:
i) e^2x = 15
To express this equation in logarithmic form, take the natural logarithm (ln) of both sides:
ln(e^2x) = ln(15)
2x = ln(15)
So, the logarithmic form is: ln(15) = 2x
ii) e^(x-4) = 1/2
Similarly, take the natural logarithm of both sides:
ln(e^(x-4)) = ln(1/2)
x - 4 = ln(1/2)
Therefore, the logarithmic form is: ln(1/2) = x - 4
iii) a^(x+1) = 5
To express this equation in logarithmic form, use the logarithm with base "a":
log(base a) [a^(x+1)] = log(base a) [5]
x + 1 = log(base a) [5]
Thus, the logarithmic form is: log(base a) [5] = x + 1
Overall, the key is to apply the logarithm function to both sides of the equation to solve for the variable.