How can I solve this log base: 2 of x +log base 4 to x = 2?
Some of the basic results used here:[math]\log_{a} b=\frac{\log_{c} b}{\log_{c} a}[/math][math]\log_{a}x=y \implies x=a^y[/math][math]\log a^b=b\log a[/math]Now focusing on the problem,[math]\log_{2} x+\log_{4} x=2[/math][math]\therefore \frac{\log x}{\log 2} +\frac{\log x}{\log 4}=2[/math][math]\therefore \frac{\log x}{\log 2} +\frac{\log x}{2\log 2}=2[/math][math]\therefore \frac{\log x}{\log 2}(1+\frac{1}{2})=2[/math][math]\therefore \frac{3}{2}×\frac{\log x}{\log 2}=2[/math][math]\therefore \log_{2} x=\frac{4}{3}[/math][math]\therefore x=2^{\frac{4}{3}}[/math][math]\therefore x=2.52[/math]Hope it helps!Note:By standard convention, here [math]\log[/math] means [math]\log_{10}[/math]. However here since we don't need to evaluate the log function, you can use any base for conversion.
How can I solve for x if log base 9 of x + log base 3 of x = 6?
Thanks for the A2A!Note that [math]\log_9{x}=\frac{1}{2}\log_3{x}[/math]. So:[math]\frac{1}{2}\log_3{x}+\log_3{x}=6\tag*{}[/math]We can simplify the LHS as:[math]\frac{3}{2}\log_3{x}=6\tag*{}[/math]Multiplying both sides by [math]\frac{2}{3}[/math]:[math]\log_3{x}=4\tag*{}[/math]Finally, raising [math]3[/math] to both sides:[math]x=3^4=9^2=81\tag*{}[/math]You can verify this by plugging it in.
How do I solve this: log 8 to the base 2 +log 8 to the base 4 + log 8 to the base 16?
[math]\log_2(8)+\log_4(8)+\log_{16}(8)[/math][math]=3+\dfrac{\ln(8)}{\ln(4)}+\dfrac{\ln(8)}{\ln(16)}[/math][math]=3+\dfrac{3\ln(2)}{2\ln(2)}+\dfrac{3\ln(2)}{4\ln(2)}[/math][math]=3+\dfrac{3}{2}+\dfrac{3}{4}[/math][math]=\dfrac{12}{4}+\dfrac{6}{4}+\dfrac{3}{4}[/math][math]=\dfrac{21}{4}[/math]
How to solve log base 2 (x+2) - log base 2 (x-5) = 3?
basically what the first guy did. but let me elaborate. log_2(x+2) - log_2(x-5)=3 -bases are the same so think of it as division log_2[(x+2)/(x-5)]=3 -now convert it into exponential form 2^3 = (x+2)/(x-5) -2 to the power of 3 is 8, and bring the denominator from the left side over to the right 8(x-5)=(x+2) expand brackets to get 8x-40=x+2 collect like terms 8x-x=2+40 7x=42 x=42/7 x=6
How do we solve the equation log (x-2) (base 3) log 3 (base x) =1?
We know that log 3 base x = log 3 / log x and log (x - 2 ) base 3 is log (x - 2) / log 3.Multiply together and ww have log (x-2) base x = 1Take exponent of both sides and we have x = x - 2Thus, there's no solution.
How do I solve this: 2 ^log 5 with base 2 (base 2 is a part of power)?
Exponent and logarithm are inverse of each other and thus, they cancel out each other. [math]a^{log_{a}(b)} \ = \ b[/math]So, [math]2^{log_{2}(5)} \ = \ 5[/math]I hope it helps!
Log-base-2^6 * log-base-6^8?
I learned how the exponents when multiplied are added and divided are subtracted but they never taught me if both logs are * and the base isn't the same, usually I solve to get them the samd but just facepalmed after 3 attempts I keep getting the answer 4 when its 3
Log(base 2)6 x log(base 6)4 =?
this question is meant to test your knowledge of the change of base formula in logs, if you recall, this formula for log(base a)b=log(any base)b/log(same base as before)a that's really hard to explain with inline text, so i assume you have been taught that so using our change of base formula we get (log(base 10)6)/(log(base 10)2) * (log(base10)4)/(log(base 10)6) the log(base 10) 6 cancels leaving (log(base 10)4)/log(base 10)2) or log(base 2)4 now remembering that a logarithm is actually the question "what exponent raised to this base will give me this number" or in this case what power of 2 is four now 2 squared is four so the answer is 2
Log (base 2) x + log (base 4) x + log (base 8) x = 11?
log(base 2)x = log(x) / log(2) log(base 4)x = log(x) / log(4) = log(x) / 2log(2) log(base 8)x = log(x) / log(8) = log(x) / 3log(2) 11 = log(2048) / log(2) log(x)/log(2) * (1 + 1/2 + 1/3) = 11 log(x) * (11/6) = 11log(2) log(x) = 6log(2) x = 2^6 = 64