Math Help - Logarithmic expression

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

Thanks in advanced.

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

Thanks in advanced.
Use $b^{\log_bx}=x$

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

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 $f(x)=b^x$ and the logarithmic function $g(x)=\log_bx$ are inverses of each other, their composites are the identity function. That is,

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

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

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

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.

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

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

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

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.