3^x=4 =7^x-1 . .. how do i solve these kind of problems?

Question

anonymous · Answer

it looks more like this $$3^{x+4}=7^{x-1}$$

anonymous · Answer

please help :(

campbell_st · Answer

ok... use logarithms 

so taking the log of both sides... doesn't matter if its base e or base 10

$$\log(3^{x +1} )= \log(7^{x -1})$$

which can be simplifed using log laws to 

$$(x + 1)\log(3) = (x - 1)\log(7)$$

distributing and collecting like terms you will get

$$Log(3) + \log(7) = x \log(7) - x \log(3)$$

factor the right hand side

$$\log(3) + \log(7) = x( \log(7)  - \log(3))$$

and divide both sides 

$$x = \frac{\log(3) + \log(7)}{\log(7)- \log(3)}$$

it can be simplified by using log laws for multiplication and addition to 

$$x = \frac{\log(21)}{\log(\frac{7}{3} )}$$

it just depends if you need an exact value answer or an answer to several decimal places..


Hope this helps