1. Logarithmic expression

Hello guys!

Could someone help me with this problem? Coz Ive been trying to solve this already for an hour but still I dont get it.

Heres the question, FInd the exact value of the given expression:

25^log58

2. Originally Posted by cjru
Hello guys!

Could someone help me with this problem? Coz Ive been trying to solve this already for an hour but still I dont get it.

Heres the question, FInd the exact value of the given expression:

25^log58

Use $\displaystyle b^{\log_bx}=x$

$\displaystyle 25^{\log_58}=5^{2\log_58}==5^{\log_58^2}=8^2=64$

3. Thank you so much.

Can you explain your step pls...

what happened to the 5^log5

Did it cancelled out?

4. Originally Posted by cjru
Thank you so much.

Can you explain your step pls...

what happened to the 5^log5

Did it cancelled out?
Since the exponential function $\displaystyle f(x)=b^x$ and the logarithmic function $\displaystyle g(x)=\log_bx$ are inverses of each other, their composites are the identity function. That is,

$\displaystyle f[g(x)]=x \ \ and \ \ g[f(x)]=x$

$\displaystyle f(\log_bx)=x \ \ and \ \ g(b^x)=x$

$\displaystyle b^{\log_bx}=x \ \ and \ \ \log_bb^x=x$

Thus, if their bases are the same, exponential and logarithmic functions "undo" each other. You can use this inverse property of exponents and logarithms to simplify expression like yours.

$\displaystyle 25^{\log_58}=5^{(2)\log_58}$

Using $\displaystyle (p)\log_bm=\log_bm^p$, we get

$\displaystyle 5^{\log_58^2}=8^2=64$

I hope I've cleared up any confusion you may have had.

5. wow! Thats makes the solution clear. Thank you so much. With your explanation I can now solve other related problems in log.

Thank you.