Logarithmic Equations

[IMIS]

Hi there. I am really struggling with logarithmic equations.

Here is my question(It is a made up question. I do not know if the numbers would work out perfectly, but I am just asking for help with understanding this type of quesion. I am not asking for anyone to solve, just trying to understand the process of solving):

2log(b)5 + .7log(b)20-log(b)20=log(b)x

So, I think b means base, correct?
So now I can write this out as.....

2log5 + .7log20-log20=x
Is this correct so far?

Now what? I know that since the base of 2log5 is 5, do I write 5^x=2+ 20^x=.7 - 20^x=1=x?
I am so confused as to what to do now. No need to solve it, just help me with understanding what to do next.

 

[Deleted member 4993]

Hi there. I am really struggling with logarithmic equations.

Here is my question(It is a made up question. I do not know if the numbers would work out perfectly, but I am just asking for help with understanding this type of quesion. I am not asking for anyone to solve, just trying to understand the process of solving):

I'll solve the problem for you - then YOU ask questions (if you did not understand any step):

2log(b)5 + .7log(b)20-log(b)20=log(b)x

the above is also written as:

2log_b5 + 0.7 * log_b20 - log_b20 = log_bx

using the rule → a*log_bc = log_bc^a

log_b(5²) + log_b(20^0.7) - log_b20 = log_bx

using the rule → log_bc + log_bd = log_b(c * d) and log_be - log_bf = log_b(e / f)

log_b(5² * 20^0.7 / 20) = log_bx

eliminating 'log' from both sides

5² * 20^0.7 / 20 = x
x = 25 / (20^0.30) = 10.17726

So, I think b means base, correct?
So now I can write this out as.....

2log5 + .7log20-log20=x →Incorrect - see above
Is this correct so far?

Now what? I know that since the base of 2log5 is 5, do I write 5^x=2+ 20^x=.7 - 20^x=1=x?
I am so confused as to what to do now. No need to solve it, just help me with understanding what to do next.

.

 

[Deleted member 4993]

Subhotosh Khan

I have a question. Where did the coefficient of 2 on the RHS of the equation come from in the second step?

Good catch - that was a mistake and fixed above.